Explanation:
Okay, so I need to figure out what (4 5/8)^55 multiplied by (8/37)^55 is. Hmm, let me start by recalling some exponent rules. When you have two numbers raised to the same exponent, you can combine them by multiplying the bases first and then raising to that exponent. So, (a^b) * (c^b) = (a*c)^b. Is that right? Let me check with smaller numbers. Like, 2^3 * 3^3 should be (2*3)^3 = 6^3 = 216. And 8*27 is 216. Yeah, that works. So, applying that here, I can combine the two bases first.

First, I need to convert the mixed number 4 5/8 into an improper fraction. Let me do that. 4 is the same as 32/8, right? Because 4*8=32. Then adding 5/8 gives me 32/8 + 5/8 = 37/8. Wait, so 4 5/8 is equal to 37/8? Let me confirm. 4 * 8 = 32, plus 5 is 37, so yes, 37/8. Okay, so the first term is (37/8)^55 multiplied by (8/37)^55.

So combining them, that would be (37/8 * 8/37)^55. Let me compute the base first. 37/8 multiplied by 8/37. The 37s cancel out, and the 8s cancel out, so that's 1. Wait, so (1)^55? Which is just 1? That seems too simple. Let me check again.

Original problem: (4 5/8)^55 * (8/37)^55. Convert 4 5/8 to 37/8. Then, (37/8)^55 * (8/37)^55. Multiplying the bases: (37/8 * 8/37) = 1. Then 1^55 is 1. So the answer is 1? That seems surprisingly simple. Let me make sure I didn't make a mistake in converting the mixed number.

4 5/8: 4 is 32/8, plus 5/8 is 37/8. Yes, that's correct. Then multiplying 37/8 by 8/37. The numerator is 37*8 and the denominator is 8*37. So yes, they are equal, so it's 1. Therefore, raising 1 to any power is still 1.

Wait, but maybe I misread the original problem? Let me check again. The first term is 4 5/8, which is 4.625, and the second term is 8/37, which is approximately 0.216. If I multiply them together, 4.625 * 0.216 ≈ 1. So, raising that product to the 55th power would be 1^55 = 1. So that seems to check out numerically.

But maybe there's a trick here? Let me think. Sometimes problems might have a typo or something, but assuming the problem is written correctly, then this seems to be the case. Let me try another approach. Let me write out the exponents:

(37/8)^55 * (8/37)^55 = [(37/8) * (8/37)]^55 = [ (37*8)/(8*37) ]^55 = [1]^55 = 1.

Yes, that's straightforward. So unless there's some hidden detail or a different interpretation of the original problem, this seems correct.

Wait, let me check if the exponents are both 55. The problem says (4 5/8)^55 multiplied by (8/37)^55. Yes, both exponents are 55. So combining them is valid.

Alternatively, if the exponents were different, we couldn't combine them like that. But since they are the same, it's okay.

Another way to think about it: Let me write both terms with exponents.

First term: (37/8)^55. Second term: (8/37)^55. Notice that (8/37) is the reciprocal of (37/8). So, (8/37) = 1/(37/8). Therefore, (8/37)^55 = [1/(37/8)]^55 = 1^55 / (37/8)^55. Therefore, multiplying by (37/8)^55 gives [ (37/8)^55 ] * [ 1 / (37/8)^55 ] = 1.

Yes, that also works. So regardless of the approach, it seems that the answer is 1.

I think that's solid. Maybe I was overcomplicating it at first, but all the different methods lead to the same conclusion. So I think the answer is 1.

**Final Answer**
\boxed{1}

To evaluate the expression:

$$
\left(4\dfrac{5}{8}\right)^{55} \cdot \left(\dfrac{8}{37}\right)^{55}
$$

we begin by simplifying the mixed number $4\dfrac{5}{8}$ into an improper fraction.

---

### Step 1: Convert the Mixed Number

$$
4\dfrac{5}{8} = \dfrac{4 \cdot 8 + 5}{8} = \dfrac{32 + 5}{8} = \dfrac{37}{8}
$$

So the expression becomes:

$$
\left(\dfrac{37}{8}\right)^{55} \cdot \left(\dfrac{8}{37}\right)^{55}
$$

---

### Step 2: Apply the Exponent Rule

Recall the exponent rule:

$$
(a^b) \cdot (c^b) = (a \cdot c)^b
$$

Applying this to our expression:

$$
\left(\dfrac{37}{8} \cdot \dfrac{8}{37}\right)^{55}
$$

---

### Step 3: Simplify the Base

Now simplify the product inside the parentheses:

$$
\dfrac{37}{8} \cdot \dfrac{8}{37} = \dfrac{37 \cdot 8}{8 \cdot 37} = \dfrac{296}{296} = 1
$$

So the entire expression becomes:

$$
1^{55}
$$

---

### Step 4: Final Simplification

$$
1^{55} = 1
$$

---

### Final Answer

$$
\boxed{1}
$$

[Assistant]
Explanation:
Okay, so I need to figure out what (4 5/8)^55 multiplied by (8/37)^55 is. Hmm, let me start by recalling some exponent rules. When you have two numbers raised to the same exponent, you can combine them by multiplying the bases first and then raising to that exponent. So, (a^b) * (c^b) = (a*c)^b. Is that right? Let me check with smaller numbers. Like, 2^3 * 3^3 should be (2*3)^3 = 6^3 = 216. And 8*27 is 216. Yeah, that works. So, applying that here, I can combine the two bases first.

First, I need to convert the mixed number 4 5/8 into an improper fraction. Let me do that. 4 is the same as 32/8, right? Because 4*8=32. Then adding 5/8 gives me 32/8 + 5/8 = 37/8. Wait, so 4 5/8 is equal to 37/8? Let me confirm. 4 * 8 = 32, plus 5 is 37, so yes, 37/8. Okay, so the first term is (37/8)^55 multiplied by (8/37)^55.

So combining them, that would be (37/8 * 8/37)^55. Let me compute the base first. 37/8 multiplied by 8/37. The 37s cancel out, and the 8s cancel out, so that's 1. Wait, so (1)^55? Which is just 1? That seems too simple. Let me check again.

Original problem: (4 5/8)^55 * (8/37)^55. Convert 4 5/8 to 37/8. Then, (37/8)^55 * (8/37)^55. Multiplying the bases: (37/8 * 8/37) = 1. Then 1^55 is 1. So the answer is 1? That seems surprisingly simple. Let me make sure I didn't make a mistake in converting the mixed number.

4 5/8: 4 is 32/8, plus 5/8 is 37/8. Yes, that's correct. Then multiplying 37/8 by 8/37. The numerator is 37*8 and the denominator is 8*37. So yes, they are equal, so it's 1. Therefore, raising 1 to any power is still 1.

Wait, but maybe I misread the original problem? Let me check again. The first term is 4 5/8, which is 4.625, and the second term is 8/37, which is approximately 0.216. If I multiply them together, 4.625 * 0.216 ≈ 1. So, raising that product to the 55th power would be 1^55 = 1. So that seems to check out numerically.

But maybe there's a trick here? Let me think. Sometimes problems might have a typo or something, but assuming the problem is written correctly, then this seems to be the case. Let me try another approach. Let me write out the exponents:

(37/8)^55 * (8/37)^55 = [(37/8) * (8/37)]^55 = [ (37*8)/(8*37) ]^55 = [1]^55 = 1.

Yes, that's straightforward. So unless there's some hidden detail or a different interpretation of the original problem, this seems correct.

Wait, let me check if the exponents are both 55. The problem says (4 5/8)^55 multiplied by (8/37)^55. Yes, both exponents are 55. So combining them is valid.

Alternatively, if the exponents were different, we couldn't combine them like that. But since they are the same, it's okay.

Another way to think about it: Let me write both terms with exponents.

First term: (37/8)^55. Second term: (8/37)^55. Notice that (8/37) is the reciprocal of (37/8). So, (8/37) = 1/(37/8). Therefore, (8/37)^55 = [1/(37/8)]^55 = 1^55 / (37/8)^55. Therefore, multiplying by (37/8)^55 gives [ (37/8)^55 ] * [ 1 / (37/8)^55 ] = 1.

Yes, that also works. So regardless of the approach, it seems that the answer is 1.

I think that's solid. Maybe I was overcomplicating it at first, but all the different methods lead to the same conclusion. So I think the answer is 1.

**Final Answer**
\boxed{1}

To evaluate the expression:

$$
\left(4\dfrac{5}{8}\right)^{55} \cdot \left(\dfrac{8}{37}\right)^{55}
$$

we begin by simplifying the mixed number $4\dfrac{5}{8}$ into an improper fraction.

---

### Step 1: Convert the Mixed Number

$$
4\dfrac{5}{8} = \dfrac{4 \cdot 8 + 5}{8} = \dfrac{32 + 5}{8} = \dfrac{37}{8}
$$

So the expression becomes:

$$
\left(\dfrac{37}{8}\right)^{55} \cdot \left(\dfrac{8}{37}\right)^{55}
$$

---

### Step 2: Apply the Exponent Rule

Recall the exponent rule:

$$
(a^b) \cdot (c^b) = (a \cdot c)^b
$$

Applying this to our expression:

$$
\left(\dfrac{37}{8} \cdot \dfrac{8}{37}\right)^{55}
$$

---

### Step 3: Simplify the Base

Now simplify the product inside the parentheses:

$$
\dfrac{37}{8} \cdot \dfrac{8}{37} = \dfrac{37 \cdot 8}{8 \cdot 37} = \dfrac{296}{296} = 1
$$

So the entire expression becomes:

$$
1^{55}
$$

---

### Step 4: Final Simplification

$$
1^{55} = 1
$$

---

### Final Answer

$$
\boxed{1}
$$

Solution:
First we convert $4\dfrac{5}{8}$ into an improper fraction: \[4\dfrac{5}{8} = 4 + \dfrac{5}{8} = \dfrac{32}{8} + \dfrac{5}{8} = \dfrac{37}{8}.\]We discover that $4\dfrac{5}{8}$ and $\dfrac{8}{37}$ are in fact reciprocals of each other. Using the fact that $(ab)^n = a^nb^n$, we get our answer: \[
\left(4\dfrac{5}{8}\right)^{55} \cdot \left(\dfrac{8}{37}\right)^{55} = \left(4\dfrac{5}{8} \cdot \dfrac{8}{37}\right)^{55} = 1^{55} = \boxed{1}.\]


Explanation:
Okay, so I need to find the minimum value of the expression $(a + b + c + d)\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right)$ where $a, b, c, d$ are positive real numbers. Hmm, let me think about how to approach this.

First, I remember that inequalities like Cauchy-Schwarz or AM-GM might be useful here. Let me recall what those are. The Cauchy-Schwarz inequality states that for any real numbers, $(\sum u_i v_i)^2 \leq (\sum u_i^2)(\sum v_i^2)$. But I'm not sure if that's directly applicable here. Maybe the AM-GM inequality? The AM-GM inequality says that the arithmetic mean is greater than or equal to the geometric mean. For positive numbers, that is.

Alternatively, I remember that for two positive numbers, the product $(x + y)(\frac{1}{x} + \frac{1}{y})$ has a minimum value of 4, achieved when $x = y$. Maybe this problem is a generalization of that? If I consider the case where all variables are equal, maybe that gives the minimum?

Let me test that idea. Suppose $a = b = c = d = k$, where $k$ is a positive real number. Then the expression becomes:

$$
(4k)\left( \frac{4}{k} \right) = 4k \times \frac{4}{k} = 16
$$

So, if all variables are equal, the value is 16. Is this the minimum? Maybe, but I need to check if there's a possibility of getting a lower value if the variables aren't equal.

Let me try with different values. Suppose I take $a = b = c = 1$ and $d$ approaching 0. Then the first part of the product, $a + b + c + d$, approaches 3 + 0 = 3. The second part, $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}$, approaches 1 + 1 + 1 + infinity = infinity. So the product approaches infinity. Similarly, if one variable becomes very large, then the first part becomes large and the second part becomes small, but I think the product might still go to infinity? Let me check.

Suppose $a$ approaches infinity, and $b = c = d = 1$. Then the first part is $a + 1 + 1 + 1 = a + 3$, which approaches infinity. The second part is $\frac{1}{a} + 1 + 1 + 1 = \frac{1}{a} + 3$, which approaches 3. So the product approaches infinity * 3 = infinity. So making variables very large or very small seems to make the product larger. So maybe the minimum occurs when all variables are equal?

But wait, maybe there's a case where some variables are different but the product is less than 16? Let me try with two variables equal and the other two different. For example, let me take $a = b = 1$ and $c = d = t$, where $t$ is some positive number. Then compute the expression.

First, compute the sum $a + b + c + d = 1 + 1 + t + t = 2 + 2t$.

Then the reciprocal sum is $\frac{1}{1} + \frac{1}{1} + \frac{1}{t} + \frac{1}{t} = 2 + \frac{2}{t}$.

So the product is $(2 + 2t)\left(2 + \frac{2}{t}\right)$. Let me compute that:

Multiply out:

$2*2 + 2*\frac{2}{t} + 2t*2 + 2t*\frac{2}{t}$

Which is $4 + \frac{4}{t} + 4t + 4$.

Simplify: $8 + 4t + \frac{4}{t}$.

Now, to find the minimum of this expression with respect to t > 0. Let me set $f(t) = 8 + 4t + \frac{4}{t}$. Take derivative: $f'(t) = 4 - \frac{4}{t^2}$. Set derivative to zero:

$4 - \frac{4}{t^2} = 0$ => $4 = \frac{4}{t^2}$ => $t^2 = 1$ => $t = 1$ (since t > 0). So minimum occurs at t = 1, which gives f(1) = 8 + 4*1 + 4/1 = 8 + 4 + 4 = 16. So even in this case, the minimum is 16. So even when we have two variables equal and the other two equal, the minimum is still 16. So maybe 16 is indeed the minimum?

But let me try another case. Suppose three variables are equal and one is different. Let me take $a = b = c = 1$ and $d = t$. Then compute the expression.

Sum: $1 + 1 + 1 + t = 3 + t$.

Reciprocal sum: $1 + 1 + 1 + 1/t = 3 + 1/t$.

Product: $(3 + t)(3 + 1/t)$. Let me expand this:

$3*3 + 3*(1/t) + t*3 + t*(1/t) = 9 + 3/t + 3t + 1 = 10 + 3t + 3/t$.

So the function is $f(t) = 10 + 3t + 3/t$. Take derivative: $f'(t) = 3 - 3/t^2$. Set to zero: $3 - 3/t^2 = 0$ => $t^2 = 1$ => t = 1. Then f(1) = 10 + 3 + 3 = 16. Again, same minimum. So even when three variables are equal and one is different, the minimum is still 16. So maybe regardless of how we set variables, the minimum is 16?

Wait, but maybe if we take different numbers? Let me try with two variables different. For example, let me take $a = 2$, $b = 2$, $c = 1$, $d = 1$. Then compute the product.

Sum: 2 + 2 + 1 + 1 = 6.

Reciprocal sum: 1/2 + 1/2 + 1 + 1 = 1 + 2 = 3.

Product: 6 * 3 = 18, which is larger than 16. So worse.

Another example: take $a = 1$, $b = 2$, $c = 3$, $d = 6$. Let me compute the sum: 1 + 2 + 3 + 6 = 12. Reciprocal sum: 1 + 1/2 + 1/3 + 1/6 = 1 + 0.5 + 0.333... + 0.166... ≈ 2. So product ≈ 12 * 2 = 24. Still higher.

Alternatively, take $a = 1$, $b = 1$, $c = 2$, $d = 2$. Then sum is 1 + 1 + 2 + 2 = 6. Reciprocal sum is 1 + 1 + 0.5 + 0.5 = 3. Product is 6 * 3 = 18. Still higher than 16.

Alternatively, take $a = 1$, $b = 1$, $c = 1$, $d = 1$. Then product is 4 * 4 = 16. So that's the same as before.

Wait, so maybe 16 is indeed the minimum? But how can I be sure? Maybe I need to use some inequality to prove that the expression is always at least 16?

Let me think about Cauchy-Schwarz. The Cauchy-Schwarz inequality in the form $(\sum u_i v_i)^2 \leq (\sum u_i^2)(\sum v_i^2)$. But how can I apply it here?

Alternatively, there's the Cauchy-Schwarz inequality in the form $(\sum x_i^2)(\sum y_i^2) \geq (\sum x_i y_i)^2$. If I set $x_i = \sqrt{a}$, $y_i = 1/\sqrt{a}$, but maybe not. Wait, let me think.

Alternatively, consider vectors $(\sqrt{a}, \sqrt{b}, \sqrt{c}, \sqrt{d})$ and $(1/\sqrt{a}, 1/\sqrt{b}, 1/\sqrt{c}, 1/\sqrt{d})$. Then their dot product is $1 + 1 + 1 + 1 = 4$. Then by Cauchy-Schwarz:

$$
(\sqrt{a} \cdot \frac{1}{\sqrt{a}} + \sqrt{b} \cdot \frac{1}{\sqrt{b}} + \sqrt{c} \cdot \frac{1}{\sqrt{c}} + \sqrt{d} \cdot \frac{1}{\sqrt{d}})^2 \leq (a + b + c + d)\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right)
$$

Which simplifies to:

$$
(1 + 1 + 1 + 1)^2 \leq (a + b + c + d)\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right)
$$

So:

$$
16 \leq (a + b + c + d)\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right)
$$

Therefore, the minimum value is 16, achieved when the vectors are proportional. That is, when $\sqrt{a}/(1/\sqrt{a}) = \sqrt{b}/(1/\sqrt{b}) = \sqrt{c}/(1/\sqrt{c}) = \sqrt{d}/(1/\sqrt{d})$, which simplifies to $a = b = c = d$. Therefore, equality holds when all variables are equal. Therefore, the minimum value is 16.

Wait, so that's a solid proof using Cauchy-Schwarz. Therefore, regardless of the values of a, b, c, d (as long as they are positive), the product is at least 16, and equality occurs when all variables are equal. Therefore, the answer is 16.

But let me check if there's another way to think about it, maybe using AM-GM. Let me try that approach as well.

The AM-GM inequality states that for non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. For four variables, the AM-GM inequality would be:

$$
\frac{a + b + c + d}{4} \geq \sqrt[4]{abcd}
$$

Similarly, for the reciprocals:

$$
\frac{\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}}{4} \geq \sqrt[4]{\frac{1}{abcd}}
$$

Multiplying these two inequalities:

$$
\frac{a + b + c + d}{4} \times \frac{\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}}{4} \geq \sqrt[4]{abcd} \times \sqrt[4]{\frac{1}{abcd}} = \sqrt[4]{1} = 1
$$

Therefore:

$$
(a + b + c + d)\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right) \geq 16
$$

So again, we get that the product is at least 16, with equality when $a = b = c = d$. Therefore, this confirms the result from Cauchy-Schwarz. So regardless of the method, the minimum is 16.

But wait, let me check if equality conditions hold. For AM-GM, equality holds when all the numbers are equal. So for the first AM-GM, equality holds when $a = b = c = d$, and for the second AM-GM, equality also holds when $1/a = 1/b = 1/c = 1/d$, which again implies $a = b = c = d$. Therefore, equality occurs when all variables are equal, so the minimum is indeed 16.

Therefore, the answer is 16. But let me think again if there's a case where variables are not equal but the product is still 16? For example, suppose two variables are equal and the other two are equal but different from the first pair. Wait, earlier when I tried with two pairs, the minimum was still 16. So even if variables are not all equal, but grouped into equal pairs, the minimum is still 16. So that suggests that the minimum is achieved not only when all variables are equal but also when they are grouped in some way. However, in those cases, when variables are grouped, the minimum is still 16. Therefore, regardless of the configuration, the minimum is 16.

Alternatively, maybe there's a way to use the Cauchy-Schwarz inequality more directly. Let me recall that for any positive real numbers, $(\sum x_i)(\sum y_i) \geq (\sum \sqrt{x_i y_i})^2$. Wait, is that true? Let me check. If I set $x_i = a_i$ and $y_i = 1/a_i$, then $(\sum a_i)(\sum 1/a_i) \geq (\sum \sqrt{a_i \cdot 1/a_i})^2 = (\sum 1)^2 = 16$. So that's exactly the same as before. Therefore, this is another way to see it.

Therefore, regardless of the approach, the minimum is 16. Therefore, I can be confident that the answer is 16.

But just to make sure, let me think if there's a case where variables are not equal but the product is less than 16. Suppose I take $a = 2$, $b = 2$, $c = 2$, $d = 2$. Then the product is 8 * 2 = 16. If I take $a = 1$, $b = 1$, $c = 1$, $d = 1$, same thing. If I take $a = 1$, $b = 1$, $c = 1$, $d = 2$, then the product is (1 + 1 + 1 + 2)(1 + 1 + 1 + 1/2) = 5 * 3.5 = 17.5, which is more than 16. If I take $a = 1$, $b = 1$, $c = 2$, $d = 2$, then the product is (1 + 1 + 2 + 2)(1 + 1 + 1/2 + 1/2) = 6 * 3 = 18. Still higher. If I take $a = 1$, $b = 2$, $c = 3$, $d = 6$, as before, the product is 12 * 2 = 24. So all these examples give higher than 16. Therefore, seems like 16 is indeed the minimum.

Alternatively, maybe take some other numbers. Let me try $a = 1$, $b = 2$, $c = 3$, $d = 6$. Wait, already did that. How about $a = 1$, $b = 1$, $c = 2$, $d = 0.5$. Then sum is 1 + 1 + 2 + 0.5 = 4.5. Reciprocal sum is 1 + 1 + 0.5 + 2 = 4.5. Then product is 4.5 * 4.5 = 20.25, which is still higher than 16. So even if I take some variables larger and some smaller, the product is higher. Therefore, seems like 16 is the minimum.

Therefore, I think I can conclude that the minimum value is 16, achieved when all variables are equal. Therefore, the answer is $\boxed{16}$.

**Final Answer**
\boxed{16}

To find the minimum value of the expression

$$
(a + b + c + d)\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right),
$$

where $a, b, c, d$ are positive real numbers, we can approach the problem using **inequalities** such as the **Cauchy-Schwarz Inequality** or the **AM-GM Inequality**.

---

### Step 1: Apply the Cauchy-Schwarz Inequality

We consider the vectors:

$$
\vec{u} = (\sqrt{a}, \sqrt{b}, \sqrt{c}, \sqrt{d}), \quad \vec{v} = \left( \frac{1}{\sqrt{a}}, \frac{1}{\sqrt{b}}, \frac{1}{\sqrt{c}}, \frac{1}{\sqrt{d}} \right).
$$

By the **Cauchy-Schwarz Inequality**, we have:

$$
(\vec{u} \cdot \vec{v})^2 \leq (\vec{u} \cdot \vec{u})(\vec{v} \cdot \vec{v}).
$$

Compute the dot products:

- $\vec{u} \cdot \vec{v} = \sqrt{a} \cdot \frac{1}{\sqrt{a}} + \sqrt{b} \cdot \frac{1}{\sqrt{b}} + \sqrt{c} \cdot \frac{1}{\sqrt{c}} + \sqrt{d} \cdot \frac{1}{\sqrt{d}} = 1 + 1 + 1 + 1 = 4$,
- $\vec{u} \cdot \vec{u} = a + b + c + d$,
- $\vec{v} \cdot \vec{v} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}$.

So the inequality becomes:

$$
4^2 \leq (a + b + c + d)\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right),
$$

which simplifies to:

$$
16 \leq (a + b + c + d)\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right).
$$

---

### Step 2: Equality Condition

Equality in the Cauchy-Schwarz inequality holds when the vectors are **proportional**, i.e.,

$$
\frac{\sqrt{a}}{1/\sqrt{a}} = \frac{\sqrt{b}}{1/\sqrt{b}} = \frac{\sqrt{c}}{1/\sqrt{c}} = \frac{\sqrt{d}}{1/\sqrt{d}}.
$$

This simplifies to:

$$
a = b = c = d.
$$

So, the minimum value of the expression is achieved when all variables are equal.

---

### Step 3: Compute the Minimum Value

Let $a = b = c = d = k$, where $k > 0$. Then:

$$
(a + b + c + d) = 4k, \quad \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right) = \frac{4}{k}.
$$

So the product becomes:

$$
(4k)\left( \frac{4}{k} \right) = 16.
$$

---

### Step 4: Confirm with AM-GM Inequality

We can also use the **AM-GM Inequality** to confirm the result. For positive real numbers $a, b, c, d$, we have:

$$
\frac{a + b + c + d}{4} \geq \sqrt[4]{abcd}, \quad \frac{\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}}{4} \geq \sqrt[4]{\frac{1}{abcd}}.
$$

Multiplying these two inequalities:

$$
\frac{a + b + c + d}{4} \cdot \frac{\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}}{4} \geq \sqrt[4]{abcd} \cdot \sqrt[4]{\frac{1}{abcd}} = 1.
$$

Multiplying both sides by 16:

$$
(a + b + c + d)\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right) \geq 16.
$$

Equality holds when $a = b = c = d$, confirming the earlier result.

---

### Final Conclusion

The minimum value of the expression is achieved when all variables are equal, and the value is:

$$
\boxed{16}
$$

[Assistant]
Explanation:
Okay, so I need to find the minimum value of the expression $(a + b + c + d)\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right)$ where $a, b, c, d$ are positive real numbers. Hmm, let me think about how to approach this.

First, I remember that inequalities like Cauchy-Schwarz or AM-GM might be useful here. Let me recall what those are. The Cauchy-Schwarz inequality states that for any real numbers, $(\sum u_i v_i)^2 \leq (\sum u_i^2)(\sum v_i^2)$. But I'm not sure if that's directly applicable here. Maybe the AM-GM inequality? The AM-GM inequality says that the arithmetic mean is greater than or equal to the geometric mean. For positive numbers, that is.

Alternatively, I remember that for two positive numbers, the product $(x + y)(\frac{1}{x} + \frac{1}{y})$ has a minimum value of 4, achieved when $x = y$. Maybe this problem is a generalization of that? If I consider the case where all variables are equal, maybe that gives the minimum?

Let me test that idea. Suppose $a = b = c = d = k$, where $k$ is a positive real number. Then the expression becomes:

$$
(4k)\left( \frac{4}{k} \right) = 4k \times \frac{4}{k} = 16
$$

So, if all variables are equal, the value is 16. Is this the minimum? Maybe, but I need to check if there's a possibility of getting a lower value if the variables aren't equal.

Let me try with different values. Suppose I take $a = b = c = 1$ and $d$ approaching 0. Then the first part of the product, $a + b + c + d$, approaches 3 + 0 = 3. The second part, $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}$, approaches 1 + 1 + 1 + infinity = infinity. So the product approaches infinity. Similarly, if one variable becomes very large, then the first part becomes large and the second part becomes small, but I think the product might still go to infinity? Let me check.

Suppose $a$ approaches infinity, and $b = c = d = 1$. Then the first part is $a + 1 + 1 + 1 = a + 3$, which approaches infinity. The second part is $\frac{1}{a} + 1 + 1 + 1 = \frac{1}{a} + 3$, which approaches 3. So the product approaches infinity * 3 = infinity. So making variables very large or very small seems to make the product larger. So maybe the minimum occurs when all variables are equal?

But wait, maybe there's a case where some variables are different but the product is less than 16? Let me try with two variables equal and the other two different. For example, let me take $a = b = 1$ and $c = d = t$, where $t$ is some positive number. Then compute the expression.

First, compute the sum $a + b + c + d = 1 + 1 + t + t = 2 + 2t$.

Then the reciprocal sum is $\frac{1}{1} + \frac{1}{1} + \frac{1}{t} + \frac{1}{t} = 2 + \frac{2}{t}$.

So the product is $(2 + 2t)\left(2 + \frac{2}{t}\right)$. Let me compute that:

Multiply out:

$2*2 + 2*\frac{2}{t} + 2t*2 + 2t*\frac{2}{t}$

Which is $4 + \frac{4}{t} + 4t + 4$.

Simplify: $8 + 4t + \frac{4}{t}$.

Now, to find the minimum of this expression with respect to t > 0. Let me set $f(t) = 8 + 4t + \frac{4}{t}$. Take derivative: $f'(t) = 4 - \frac{4}{t^2}$. Set derivative to zero:

$4 - \frac{4}{t^2} = 0$ => $4 = \frac{4}{t^2}$ => $t^2 = 1$ => $t = 1$ (since t > 0). So minimum occurs at t = 1, which gives f(1) = 8 + 4*1 + 4/1 = 8 + 4 + 4 = 16. So even in this case, the minimum is 16. So even when we have two variables equal and the other two equal, the minimum is still 16. So maybe 16 is indeed the minimum?

But let me try another case. Suppose three variables are equal and one is different. Let me take $a = b = c = 1$ and $d = t$. Then compute the expression.

Sum: $1 + 1 + 1 + t = 3 + t$.

Reciprocal sum: $1 + 1 + 1 + 1/t = 3 + 1/t$.

Product: $(3 + t)(3 + 1/t)$. Let me expand this:

$3*3 + 3*(1/t) + t*3 + t*(1/t) = 9 + 3/t + 3t + 1 = 10 + 3t + 3/t$.

So the function is $f(t) = 10 + 3t + 3/t$. Take derivative: $f'(t) = 3 - 3/t^2$. Set to zero: $3 - 3/t^2 = 0$ => $t^2 = 1$ => t = 1. Then f(1) = 10 + 3 + 3 = 16. Again, same minimum. So even when three variables are equal and one is different, the minimum is still 16. So maybe regardless of how we set variables, the minimum is 16?

Wait, but maybe if we take different numbers? Let me try with two variables different. For example, let me take $a = 2$, $b = 2$, $c = 1$, $d = 1$. Then compute the product.

Sum: 2 + 2 + 1 + 1 = 6.

Reciprocal sum: 1/2 + 1/2 + 1 + 1 = 1 + 2 = 3.

Product: 6 * 3 = 18, which is larger than 16. So worse.

Another example: take $a = 1$, $b = 2$, $c = 3$, $d = 6$. Let me compute the sum: 1 + 2 + 3 + 6 = 12. Reciprocal sum: 1 + 1/2 + 1/3 + 1/6 = 1 + 0.5 + 0.333... + 0.166... ≈ 2. So product ≈ 12 * 2 = 24. Still higher.

Alternatively, take $a = 1$, $b = 1$, $c = 2$, $d = 2$. Then sum is 1 + 1 + 2 + 2 = 6. Reciprocal sum is 1 + 1 + 0.5 + 0.5 = 3. Product is 6 * 3 = 18. Still higher than 16.

Alternatively, take $a = 1$, $b = 1$, $c = 1$, $d = 1$. Then product is 4 * 4 = 16. So that's the same as before.

Wait, so maybe 16 is indeed the minimum? But how can I be sure? Maybe I need to use some inequality to prove that the expression is always at least 16?

Let me think about Cauchy-Schwarz. The Cauchy-Schwarz inequality in the form $(\sum u_i v_i)^2 \leq (\sum u_i^2)(\sum v_i^2)$. But how can I apply it here?

Alternatively, there's the Cauchy-Schwarz inequality in the form $(\sum x_i^2)(\sum y_i^2) \geq (\sum x_i y_i)^2$. If I set $x_i = \sqrt{a}$, $y_i = 1/\sqrt{a}$, but maybe not. Wait, let me think.

Alternatively, consider vectors $(\sqrt{a}, \sqrt{b}, \sqrt{c}, \sqrt{d})$ and $(1/\sqrt{a}, 1/\sqrt{b}, 1/\sqrt{c}, 1/\sqrt{d})$. Then their dot product is $1 + 1 + 1 + 1 = 4$. Then by Cauchy-Schwarz:

$$
(\sqrt{a} \cdot \frac{1}{\sqrt{a}} + \sqrt{b} \cdot \frac{1}{\sqrt{b}} + \sqrt{c} \cdot \frac{1}{\sqrt{c}} + \sqrt{d} \cdot \frac{1}{\sqrt{d}})^2 \leq (a + b + c + d)\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right)
$$

Which simplifies to:

$$
(1 + 1 + 1 + 1)^2 \leq (a + b + c + d)\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right)
$$

So:

$$
16 \leq (a + b + c + d)\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right)
$$

Therefore, the minimum value is 16, achieved when the vectors are proportional. That is, when $\sqrt{a}/(1/\sqrt{a}) = \sqrt{b}/(1/\sqrt{b}) = \sqrt{c}/(1/\sqrt{c}) = \sqrt{d}/(1/\sqrt{d})$, which simplifies to $a = b = c = d$. Therefore, equality holds when all variables are equal. Therefore, the minimum value is 16.

Wait, so that's a solid proof using Cauchy-Schwarz. Therefore, regardless of the values of a, b, c, d (as long as they are positive), the product is at least 16, and equality occurs when all variables are equal. Therefore, the answer is 16.

But let me check if there's another way to think about it, maybe using AM-GM. Let me try that approach as well.

The AM-GM inequality states that for non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. For four variables, the AM-GM inequality would be:

$$
\frac{a + b + c + d}{4} \geq \sqrt[4]{abcd}
$$

Similarly, for the reciprocals:

$$
\frac{\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}}{4} \geq \sqrt[4]{\frac{1}{abcd}}
$$

Multiplying these two inequalities:

$$
\frac{a + b + c + d}{4} \times \frac{\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}}{4} \geq \sqrt[4]{abcd} \times \sqrt[4]{\frac{1}{abcd}} = \sqrt[4]{1} = 1
$$

Therefore:

$$
(a + b + c + d)\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right) \geq 16
$$

So again, we get that the product is at least 16, with equality when $a = b = c = d$. Therefore, this confirms the result from Cauchy-Schwarz. So regardless of the method, the minimum is 16.

But wait, let me check if equality conditions hold. For AM-GM, equality holds when all the numbers are equal. So for the first AM-GM, equality holds when $a = b = c = d$, and for the second AM-GM, equality also holds when $1/a = 1/b = 1/c = 1/d$, which again implies $a = b = c = d$. Therefore, equality occurs when all variables are equal, so the minimum is indeed 16.

Therefore, the answer is 16. But let me think again if there's a case where variables are not equal but the product is still 16? For example, suppose two variables are equal and the other two are equal but different from the first pair. Wait, earlier when I tried with two pairs, the minimum was still 16. So even if variables are not all equal, but grouped into equal pairs, the minimum is still 16. So that suggests that the minimum is achieved not only when all variables are equal but also when they are grouped in some way. However, in those cases, when variables are grouped, the minimum is still 16. Therefore, regardless of the configuration, the minimum is 16.

Alternatively, maybe there's a way to use the Cauchy-Schwarz inequality more directly. Let me recall that for any positive real numbers, $(\sum x_i)(\sum y_i) \geq (\sum \sqrt{x_i y_i})^2$. Wait, is that true? Let me check. If I set $x_i = a_i$ and $y_i = 1/a_i$, then $(\sum a_i)(\sum 1/a_i) \geq (\sum \sqrt{a_i \cdot 1/a_i})^2 = (\sum 1)^2 = 16$. So that's exactly the same as before. Therefore, this is another way to see it.

Therefore, regardless of the approach, the minimum is 16. Therefore, I can be confident that the answer is 16.

But just to make sure, let me think if there's a case where variables are not equal but the product is less than 16. Suppose I take $a = 2$, $b = 2$, $c = 2$, $d = 2$. Then the product is 8 * 2 = 16. If I take $a = 1$, $b = 1$, $c = 1$, $d = 1$, same thing. If I take $a = 1$, $b = 1$, $c = 1$, $d = 2$, then the product is (1 + 1 + 1 + 2)(1 + 1 + 1 + 1/2) = 5 * 3.5 = 17.5, which is more than 16. If I take $a = 1$, $b = 1$, $c = 2$, $d = 2$, then the product is (1 + 1 + 2 + 2)(1 + 1 + 1/2 + 1/2) = 6 * 3 = 18. Still higher. If I take $a = 1$, $b = 2$, $c = 3$, $d = 6$, as before, the product is 12 * 2 = 24. So all these examples give higher than 16. Therefore, seems like 16 is indeed the minimum.

Alternatively, maybe take some other numbers. Let me try $a = 1$, $b = 2$, $c = 3$, $d = 6$. Wait, already did that. How about $a = 1$, $b = 1$, $c = 2$, $d = 0.5$. Then sum is 1 + 1 + 2 + 0.5 = 4.5. Reciprocal sum is 1 + 1 + 0.5 + 2 = 4.5. Then product is 4.5 * 4.5 = 20.25, which is still higher than 16. So even if I take some variables larger and some smaller, the product is higher. Therefore, seems like 16 is the minimum.

Therefore, I think I can conclude that the minimum value is 16, achieved when all variables are equal. Therefore, the answer is $\boxed{16}$.

**Final Answer**
\boxed{16}

To find the minimum value of the expression

$$
(a + b + c + d)\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right),
$$

where $a, b, c, d$ are positive real numbers, we can approach the problem using **inequalities** such as the **Cauchy-Schwarz Inequality** or the **AM-GM Inequality**.

---

### Step 1: Apply the Cauchy-Schwarz Inequality

We consider the vectors:

$$
\vec{u} = (\sqrt{a}, \sqrt{b}, \sqrt{c}, \sqrt{d}), \quad \vec{v} = \left( \frac{1}{\sqrt{a}}, \frac{1}{\sqrt{b}}, \frac{1}{\sqrt{c}}, \frac{1}{\sqrt{d}} \right).
$$

By the **Cauchy-Schwarz Inequality**, we have:

$$
(\vec{u} \cdot \vec{v})^2 \leq (\vec{u} \cdot \vec{u})(\vec{v} \cdot \vec{v}).
$$

Compute the dot products:

- $\vec{u} \cdot \vec{v} = \sqrt{a} \cdot \frac{1}{\sqrt{a}} + \sqrt{b} \cdot \frac{1}{\sqrt{b}} + \sqrt{c} \cdot \frac{1}{\sqrt{c}} + \sqrt{d} \cdot \frac{1}{\sqrt{d}} = 1 + 1 + 1 + 1 = 4$,
- $\vec{u} \cdot \vec{u} = a + b + c + d$,
- $\vec{v} \cdot \vec{v} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}$.

So the inequality becomes:

$$
4^2 \leq (a + b + c + d)\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right),
$$

which simplifies to:

$$
16 \leq (a + b + c + d)\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right).
$$

---

### Step 2: Equality Condition

Equality in the Cauchy-Schwarz inequality holds when the vectors are **proportional**, i.e.,

$$
\frac{\sqrt{a}}{1/\sqrt{a}} = \frac{\sqrt{b}}{1/\sqrt{b}} = \frac{\sqrt{c}}{1/\sqrt{c}} = \frac{\sqrt{d}}{1/\sqrt{d}}.
$$

This simplifies to:

$$
a = b = c = d.
$$

So, the minimum value of the expression is achieved when all variables are equal.

---

### Step 3: Compute the Minimum Value

Let $a = b = c = d = k$, where $k > 0$. Then:

$$
(a + b + c + d) = 4k, \quad \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right) = \frac{4}{k}.
$$

So the product becomes:

$$
(4k)\left( \frac{4}{k} \right) = 16.
$$

---

### Step 4: Confirm with AM-GM Inequality

We can also use the **AM-GM Inequality** to confirm the result. For positive real numbers $a, b, c, d$, we have:

$$
\frac{a + b + c + d}{4} \geq \sqrt[4]{abcd}, \quad \frac{\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}}{4} \geq \sqrt[4]{\frac{1}{abcd}}.
$$

Multiplying these two inequalities:

$$
\frac{a + b + c + d}{4} \cdot \frac{\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}}{4} \geq \sqrt[4]{abcd} \cdot \sqrt[4]{\frac{1}{abcd}} = 1.
$$

Multiplying both sides by 16:

$$
(a + b + c + d)\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right) \geq 16.
$$

Equality holds when $a = b = c = d$, confirming the earlier result.

---

### Final Conclusion

The minimum value of the expression is achieved when all variables are equal, and the value is:

$$
\boxed{16}
$$

Solution:
By AM-GM,
\[a + b + c + d \ge 4 \sqrt[4]{abcd},\]and
\[\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \ge 4 \sqrt[4]{\frac{1}{abcd}},\]so
\[(a + b + c + d) \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right) \ge 4 \sqrt[4]{abcd} \cdot 4 \sqrt[4]{\frac{1}{abcd}} = 16.\]Equality occurs when $a = b = c = d,$ so the minimum value is $\boxed{16}.$


Explanation:
Okay, so I need to solve this equation: (6r² - 19r - 7)/(2r - 7) = 4r - 3. Hmm, let me think about how to approach this.

First, I remember that when there's a fraction equal to something, maybe I can cross-multiply or eliminate the denominator. Let me try that. If I multiply both sides by (2r - 7), that should get rid of the denominator. Let me write that down:

Multiply both sides by (2r - 7):

6r² - 19r - 7 = (4r - 3)(2r - 7)

Okay, now I need to expand the right side. Let me do that step by step.

First, multiply 4r by (2r - 7):

4r * 2r = 8r²

4r * (-7) = -28r

Then, multiply -3 by (2r - 7):

-3 * 2r = -6r

-3 * (-7) = 21

So combining all those terms:

8r² - 28r - 6r + 21

Combine like terms: -28r -6r is -34r

So the right side becomes 8r² - 34r + 21

So now the equation is:

6r² - 19r - 7 = 8r² - 34r + 21

Hmm, let me bring all terms to one side so that I can solve for r. Let me subtract the right side from both sides:

6r² - 19r - 7 - 8r² + 34r - 21 = 0

Wait, let me check that again. If I subtract 8r² -34r +21 from both sides, it's:

Left side minus right side equals zero:

(6r² - 19r -7) - (8r² -34r +21) = 0

So distribute the negative sign:

6r² -19r -7 -8r² +34r -21 = 0

Now combine like terms:

6r² -8r² = -2r²

-19r +34r = 15r

-7 -21 = -28

So the equation simplifies to:

-2r² +15r -28 = 0

Hmm, maybe I should write it as:

-2r² +15r -28 = 0

But usually, it's better to have the coefficient of r² positive. Let me multiply both sides by -1:

2r² -15r +28 = 0

Okay, so now I have a quadratic equation: 2r² -15r +28 = 0

Now, I need to solve this quadratic. Let me check if it factors. Let me think of two numbers that multiply to 2*28=56 and add up to -15. Wait, since the middle term is -15r and the constant term is +28, the numbers should multiply to +56 and add to -15. Let me think... factors of 56: 1 & 56, 2 & 28, 4 & 14, 7 & 8.

Looking for two numbers that add up to 15 (since they are both negative, their product is positive and sum is negative). Wait, the original equation is 2r² -15r +28. So the numbers should multiply to 2*28=56 and add to -15. So maybe -7 and -8? Because -7 * -8 = 56 and -7 + (-8) = -15. Yes! Perfect.

So, let me rewrite the middle term using these numbers:

2r² -7r -8r +28 = 0

Now, group the first two terms and the last two terms:

(2r² -7r) - (8r -28) = 0

Wait, hold on. Let me check the grouping again. Wait, the original is 2r² -7r -8r +28. So grouping as (2r² -7r) and (-8r +28).

Factor out common terms from each group:

From the first group, 2r² -7r, factor out an r: r(2r -7)

From the second group, -8r +28, factor out -4? Let me see. -8r +28 can be written as -4*(2r) + (-4)*(-7) = -4*(2r -7). Wait, let me check:

-4*(2r -7) = -8r +28. Yes! Perfect.

So now, the expression becomes:

r(2r -7) -4(2r -7) = 0

Now, factor out the common term (2r -7):

(2r -7)(r -4) = 0

So, the factored form is (2r -7)(r -4) = 0

Therefore, the solutions are when 2r -7 = 0 or r -4 = 0.

Solving 2r -7 = 0: 2r =7 => r=7/2

Solving r -4 =0: r=4

So, the solutions are r=7/2 and r=4.

But wait, I need to check if these solutions make the original denominator zero. The original equation had a denominator of 2r -7. So if 2r -7 =0, then r=7/2. So, r=7/2 would make the denominator zero. Therefore, r=7/2 is an extraneous solution. Therefore, we must reject it.

Therefore, the only valid solution is r=4.

Wait, let me check that again. Let me verify both solutions in the original equation to make sure.

First, check r=4:

Left side: (6*(4)^2 -19*(4) -7)/(2*(4) -7)

Calculate numerator: 6*16 = 96; 19*4=76; so 96 -76 -7 = 96 -83 = 13

Denominator: 8 -7 =1

So left side is 13/1 =13

Right side: 4*(4) -3 =16 -3=13. Okay, so 13=13. That works.

Now check r=7/2:

Left side: numerator is 6*(7/2)^2 -19*(7/2) -7

First compute (7/2)^2 = 49/4. Multiply by 6: 6*(49/4)= 294/4 = 147/2

Then, 19*(7/2) = 133/2

So numerator: 147/2 -133/2 -7 = (147 -133)/2 -7 = 14/2 -7 =7 -7=0

Denominator: 2*(7/2) -7 =7 -7=0

So left side is 0/0, which is undefined. Therefore, r=7/2 is not a valid solution. Therefore, only r=4 is the solution.

Therefore, the answer is r=4.

But let me just recap the steps to make sure I didn't skip anything.

Original equation: (6r² -19r -7)/(2r -7) = 4r -3

Multiply both sides by (2r -7): 6r² -19r -7 = (4r -3)(2r -7)

Expanded right side: 8r² -34r +21

Subtracting right side from left side: 6r² -19r -7 -8r² +34r -21 = -2r² +15r -28 =0

Multiply by -1: 2r² -15r +28=0

Factored into (2r -7)(r -4)=0, leading to r=7/2 and r=4. Then checked for extraneous solutions, found r=7/2 invalid, so only r=4 is valid.

Yes, that seems correct. So the answer is 4.

**Final Answer**
The value of $ r $ is \boxed{4}.

To solve the equation:

$$
\frac{6r^2 - 19r - 7}{2r - 7} = 4r - 3
$$

we begin by eliminating the denominator. Multiply both sides of the equation by $ 2r - 7 $, assuming $ 2r - 7 \neq 0 $ (we'll check this later):

$$
6r^2 - 19r - 7 = (4r - 3)(2r - 7)
$$

---

### Step 1: Expand the Right-Hand Side

Use the distributive property to expand $ (4r - 3)(2r - 7) $:

$$
(4r - 3)(2r - 7) = 4r \cdot 2r + 4r \cdot (-7) + (-3) \cdot 2r + (-3) \cdot (-7)
$$
$$
= 8r^2 - 28r - 6r + 21 = 8r^2 - 34r + 21
$$

So the equation becomes:

$$
6r^2 - 19r - 7 = 8r^2 - 34r + 21
$$

---

### Step 2: Move All Terms to One Side

Subtract the right-hand side from both sides:

$$
6r^2 - 19r - 7 - (8r^2 - 34r + 21) = 0
$$
$$
6r^2 - 19r - 7 - 8r^2 + 34r - 21 = 0
$$

Combine like terms:

$$
(6r^2 - 8r^2) + (-19r + 34r) + (-7 - 21) = 0
$$
$$
-2r^2 + 15r - 28 = 0
$$

Multiply both sides by $-1$ to make the leading coefficient positive:

$$
2r^2 - 15r + 28 = 0
$$

---

### Step 3: Solve the Quadratic Equation

We factor the quadratic:

$$
2r^2 - 15r + 28 = (2r - 7)(r - 4)
$$

Set each factor equal to zero:

$$
2r - 7 = 0 \Rightarrow r = \frac{7}{2}
$$
$$
r - 4 = 0 \Rightarrow r = 4
$$

---

### Step 4: Check for Extraneous Solutions

Recall that the original equation has a denominator $ 2r - 7 $. If $ r = \frac{7}{2} $, then the denominator becomes zero, which is undefined. Therefore, $ r = \frac{7}{2} $ is **not** a valid solution.

Now check $ r = 4 $ in the original equation:

**Left-hand side:**

$$
\frac{6(4)^2 - 19(4) - 7}{2(4) - 7} = \frac{96 - 76 - 7}{8 - 7} = \frac{13}{1} = 13
$$

**Right-hand side:**

$$
4(4) - 3 = 16 - 3 = 13
$$

Both sides are equal, so $ r = 4 $ is a valid solution.

---

### Final Answer

$$
\boxed{4}
$$

[Assistant]
Explanation:
Okay, so I need to solve this equation: (6r² - 19r - 7)/(2r - 7) = 4r - 3. Hmm, let me think about how to approach this.

First, I remember that when there's a fraction equal to something, maybe I can cross-multiply or eliminate the denominator. Let me try that. If I multiply both sides by (2r - 7), that should get rid of the denominator. Let me write that down:

Multiply both sides by (2r - 7):

6r² - 19r - 7 = (4r - 3)(2r - 7)

Okay, now I need to expand the right side. Let me do that step by step.

First, multiply 4r by (2r - 7):

4r * 2r = 8r²

4r * (-7) = -28r

Then, multiply -3 by (2r - 7):

-3 * 2r = -6r

-3 * (-7) = 21

So combining all those terms:

8r² - 28r - 6r + 21

Combine like terms: -28r -6r is -34r

So the right side becomes 8r² - 34r + 21

So now the equation is:

6r² - 19r - 7 = 8r² - 34r + 21

Hmm, let me bring all terms to one side so that I can solve for r. Let me subtract the right side from both sides:

6r² - 19r - 7 - 8r² + 34r - 21 = 0

Wait, let me check that again. If I subtract 8r² -34r +21 from both sides, it's:

Left side minus right side equals zero:

(6r² - 19r -7) - (8r² -34r +21) = 0

So distribute the negative sign:

6r² -19r -7 -8r² +34r -21 = 0

Now combine like terms:

6r² -8r² = -2r²

-19r +34r = 15r

-7 -21 = -28

So the equation simplifies to:

-2r² +15r -28 = 0

Hmm, maybe I should write it as:

-2r² +15r -28 = 0

But usually, it's better to have the coefficient of r² positive. Let me multiply both sides by -1:

2r² -15r +28 = 0

Okay, so now I have a quadratic equation: 2r² -15r +28 = 0

Now, I need to solve this quadratic. Let me check if it factors. Let me think of two numbers that multiply to 2*28=56 and add up to -15. Wait, since the middle term is -15r and the constant term is +28, the numbers should multiply to +56 and add to -15. Let me think... factors of 56: 1 & 56, 2 & 28, 4 & 14, 7 & 8.

Looking for two numbers that add up to 15 (since they are both negative, their product is positive and sum is negative). Wait, the original equation is 2r² -15r +28. So the numbers should multiply to 2*28=56 and add to -15. So maybe -7 and -8? Because -7 * -8 = 56 and -7 + (-8) = -15. Yes! Perfect.

So, let me rewrite the middle term using these numbers:

2r² -7r -8r +28 = 0

Now, group the first two terms and the last two terms:

(2r² -7r) - (8r -28) = 0

Wait, hold on. Let me check the grouping again. Wait, the original is 2r² -7r -8r +28. So grouping as (2r² -7r) and (-8r +28).

Factor out common terms from each group:

From the first group, 2r² -7r, factor out an r: r(2r -7)

From the second group, -8r +28, factor out -4? Let me see. -8r +28 can be written as -4*(2r) + (-4)*(-7) = -4*(2r -7). Wait, let me check:

-4*(2r -7) = -8r +28. Yes! Perfect.

So now, the expression becomes:

r(2r -7) -4(2r -7) = 0

Now, factor out the common term (2r -7):

(2r -7)(r -4) = 0

So, the factored form is (2r -7)(r -4) = 0

Therefore, the solutions are when 2r -7 = 0 or r -4 = 0.

Solving 2r -7 = 0: 2r =7 => r=7/2

Solving r -4 =0: r=4

So, the solutions are r=7/2 and r=4.

But wait, I need to check if these solutions make the original denominator zero. The original equation had a denominator of 2r -7. So if 2r -7 =0, then r=7/2. So, r=7/2 would make the denominator zero. Therefore, r=7/2 is an extraneous solution. Therefore, we must reject it.

Therefore, the only valid solution is r=4.

Wait, let me check that again. Let me verify both solutions in the original equation to make sure.

First, check r=4:

Left side: (6*(4)^2 -19*(4) -7)/(2*(4) -7)

Calculate numerator: 6*16 = 96; 19*4=76; so 96 -76 -7 = 96 -83 = 13

Denominator: 8 -7 =1

So left side is 13/1 =13

Right side: 4*(4) -3 =16 -3=13. Okay, so 13=13. That works.

Now check r=7/2:

Left side: numerator is 6*(7/2)^2 -19*(7/2) -7

First compute (7/2)^2 = 49/4. Multiply by 6: 6*(49/4)= 294/4 = 147/2

Then, 19*(7/2) = 133/2

So numerator: 147/2 -133/2 -7 = (147 -133)/2 -7 = 14/2 -7 =7 -7=0

Denominator: 2*(7/2) -7 =7 -7=0

So left side is 0/0, which is undefined. Therefore, r=7/2 is not a valid solution. Therefore, only r=4 is the solution.

Therefore, the answer is r=4.

But let me just recap the steps to make sure I didn't skip anything.

Original equation: (6r² -19r -7)/(2r -7) = 4r -3

Multiply both sides by (2r -7): 6r² -19r -7 = (4r -3)(2r -7)

Expanded right side: 8r² -34r +21

Subtracting right side from left side: 6r² -19r -7 -8r² +34r -21 = -2r² +15r -28 =0

Multiply by -1: 2r² -15r +28=0

Factored into (2r -7)(r -4)=0, leading to r=7/2 and r=4. Then checked for extraneous solutions, found r=7/2 invalid, so only r=4 is valid.

Yes, that seems correct. So the answer is 4.

**Final Answer**
The value of $ r $ is \boxed{4}.

To solve the equation:

$$
\frac{6r^2 - 19r - 7}{2r - 7} = 4r - 3
$$

we begin by eliminating the denominator. Multiply both sides of the equation by $ 2r - 7 $, assuming $ 2r - 7 \neq 0 $ (we'll check this later):

$$
6r^2 - 19r - 7 = (4r - 3)(2r - 7)
$$

---

### Step 1: Expand the Right-Hand Side

Use the distributive property to expand $ (4r - 3)(2r - 7) $:

$$
(4r - 3)(2r - 7) = 4r \cdot 2r + 4r \cdot (-7) + (-3) \cdot 2r + (-3) \cdot (-7)
$$
$$
= 8r^2 - 28r - 6r + 21 = 8r^2 - 34r + 21
$$

So the equation becomes:

$$
6r^2 - 19r - 7 = 8r^2 - 34r + 21
$$

---

### Step 2: Move All Terms to One Side

Subtract the right-hand side from both sides:

$$
6r^2 - 19r - 7 - (8r^2 - 34r + 21) = 0
$$
$$
6r^2 - 19r - 7 - 8r^2 + 34r - 21 = 0
$$

Combine like terms:

$$
(6r^2 - 8r^2) + (-19r + 34r) + (-7 - 21) = 0
$$
$$
-2r^2 + 15r - 28 = 0
$$

Multiply both sides by $-1$ to make the leading coefficient positive:

$$
2r^2 - 15r + 28 = 0
$$

---

### Step 3: Solve the Quadratic Equation

We factor the quadratic:

$$
2r^2 - 15r + 28 = (2r - 7)(r - 4)
$$

Set each factor equal to zero:

$$
2r - 7 = 0 \Rightarrow r = \frac{7}{2}
$$
$$
r - 4 = 0 \Rightarrow r = 4
$$

---

### Step 4: Check for Extraneous Solutions

Recall that the original equation has a denominator $ 2r - 7 $. If $ r = \frac{7}{2} $, then the denominator becomes zero, which is undefined. Therefore, $ r = \frac{7}{2} $ is **not** a valid solution.

Now check $ r = 4 $ in the original equation:

**Left-hand side:**

$$
\frac{6(4)^2 - 19(4) - 7}{2(4) - 7} = \frac{96 - 76 - 7}{8 - 7} = \frac{13}{1} = 13
$$

**Right-hand side:**

$$
4(4) - 3 = 16 - 3 = 13
$$

Both sides are equal, so $ r = 4 $ is a valid solution.

---

### Final Answer

$$
\boxed{4}
$$

Solution: Factoring the quadratic on the left gives $$\frac{(2r-7)(3r+1)}{2r-7} = 4r-3.$$Canceling the common factor on the left gives $3r + 1 = 4r - 3$.  Solving this equation gives $r = \boxed{4}$.