simplify.
39. $sqrt3{16y^{4}z^{12}}$
40. $sqrt3{-54x^{6}y^{11}}$
41. $\frac{x + 1}{sqrt{x}-1}$
42. $\frac{x - 2}{sqrt{x^{2}-4}}$
43. $3sqrt{24x}-2sqrt{54x}+sqrt{48}$
44. $5sqrt{18c}+3sqrt{72c}+$

Question

Accepted Answer

### Problem 39: Simplify $\boldsymbol{\sqrt[3]{16y^4z^{12}}}$
# Explanation:
## Step1: Factor the radicand
Factor $16$ as $8	imes2$, $y^4$ as $y^3	imes y$, and $z^{12}$ as $(z^4)^3$. So we have:
$$\sqrt[3]{16y^4z^{12}}=\sqrt[3]{8	imes2	imes y^3	imes y	imes (z^4)^3}$$
## Step2: Use the property of cube roots $\sqrt[3]{ab}=\sqrt[3]{a}\cdot\sqrt[3]{b}$
Separate the perfect cubes from the rest:
$$\sqrt[3]{8}\cdot\sqrt[3]{y^3}\cdot\sqrt[3]{(z^4)^3}\cdot\sqrt[3]{2y}$$
## Step3: Simplify the perfect cube roots
We know that $\sqrt[3]{8} = 2$, $\sqrt[3]{y^3}=y$, and $\sqrt[3]{(z^4)^3}=z^4$. So substituting these values in, we get:
$$2	imes y	imes z^4	imes\sqrt[3]{2y}=2yz^4\sqrt[3]{2y}$$
# Answer:
$2yz^4\sqrt[3]{2y}$

### Problem 40: Simplify $\boldsymbol{\sqrt[3]{-54x^6y^{11}}}$
# Explanation:
## Step1: Factor the radicand
Factor $-54$ as $-27	imes2$, $x^6$ as $(x^2)^3$, and $y^{11}$ as $y^9	imes y^2=(y^3)^3	imes y^2$. So we have:
$$\sqrt[3]{-54x^6y^{11}}=\sqrt[3]{-27	imes2	imes (x^2)^3	imes (y^3)^3	imes y^2}$$
## Step2: Use the property of cube roots $\sqrt[3]{ab}=\sqrt[3]{a}\cdot\sqrt[3]{b}$
Separate the perfect cubes from the rest:
$$\sqrt[3]{-27}\cdot\sqrt[3]{(x^2)^3}\cdot\sqrt[3]{(y^3)^3}\cdot\sqrt[3]{2y^2}$$
## Step3: Simplify the perfect cube roots
We know that $\sqrt[3]{-27}=-3$, $\sqrt[3]{(x^2)^3}=x^2$, and $\sqrt[3]{(y^3)^3}=y^3$. So substituting these values in, we get:
$$-3	imes x^2	imes y^3	imes\sqrt[3]{2y^2}=-3x^2y^3\sqrt[3]{2y^2}$$
# Answer:
$-3x^2y^3\sqrt[3]{2y^2}$

### Problem 41: Simplify $\boldsymbol{\frac{x + 1}{\sqrt{x}-1}}$
# Explanation:
## Step1: Recognize the numerator as a difference of squares (or use substitution)
Notice that $x + 1$ can be written as $(\sqrt{x})^2+1$? Wait, no. Wait, actually, if we multiply numerator and denominator by $\sqrt{x}+1$ (the conjugate of the denominator), we can rationalize the denominator. So:
$$\frac{x + 1}{\sqrt{x}-1}	imes\frac{\sqrt{x}+1}{\sqrt{x}+1}=\frac{(x + 1)(\sqrt{x}+1)}{(\sqrt{x})^2-1^2}$$
## Step2: Simplify the denominator
The denominator $(\sqrt{x})^2-1^2=x - 1$. So now we have:
$$\frac{(x + 1)(\sqrt{x}+1)}{x - 1}$$
Wait, but also, note that $x + 1$ and $x - 1$ don't factor further, but wait, actually, another way: Wait, $x + 1$ is not a difference of squares, but wait, maybe I made a mistake. Wait, actually, $x - 1=(\sqrt{x}-1)(\sqrt{x}+1)$, and the numerator is $x + 1$. Wait, but maybe we can also note that $x + 1$ can be written as $(\sqrt{x})^2+1$, but that doesn't help. Wait, no, actually, let's check the multiplication again. Wait, $(a - b)(a + b)=a^2 - b^2$, so denominator is $x - 1$, numerator is $(x + 1)(\sqrt{x}+1)$. But maybe there's a better way. Wait, actually, $x + 1$ can be written as $(\sqrt{x})^2+1$, but that's not a difference of squares. Wait, maybe the problem is that $x + 1$ is not factorable with the denominator, but wait, let's re - examine. Wait, maybe I made a mistake in the approach. Wait, actually, $x+1 = (\sqrt{x})^2 + 1$, but if we consider that maybe the numerator is $x + 1=(\sqrt{x})^2+1$, but that's not a difference of squares. Wait, no, wait, the original expression: $\frac{x + 1}{\sqrt{x}-1}$. Let's let $t=\sqrt{x}$, so $x = t^2$, then the expression becomes $\frac{t^2 + 1}{t - 1}$. Now, $t^2+1$ doesn't factor over the reals, but wait, that can't be. Wait, maybe I misread the numerator. Wait, the numerator is $x + 1$, denominator is $\sqrt{x}-1$. Wait, maybe there's a typo, but assuming it's correct, then the rationalized form is $\frac{(x + 1)(\sqrt{x}+1)}{x - 1}$. But wait, another way: Wait, $x+1=(\sqrt{x})^2 + 1$, but if we consider that maybe the numerator is $x - 1+2$, but that also doesn't help. Wait, no, perhaps I made a mistake in the initial step. Wait, let's do the multiplication correctly:

$$\frac{x + 1}{\sqrt{x}-1}	imes\frac{\sqrt{x}+1}{\sqrt{x}+1}=\frac{(x + 1)(\sqrt{x}+1)}{x - 1}=\frac{x\sqrt{x}+x+\sqrt{x}+1}{x - 1}$$

But that's not simpler. Wait, maybe the numerator was supposed to be $x - 1$? If it was $x - 1$, then it would be $\frac{(x - 1)(\sqrt{x}+1)}{x - 1}=\sqrt{x}+1$ (for $x
eq1$). But since the numerator is $x + 1$, we have to go with the rationalized form. Wait, but maybe I made a mistake. Wait, let's check with $x = 4$. Then the original expression: $\frac{4 + 1}{\sqrt{4}-1}=\frac{5}{2 - 1}=5$. The rationalized form: $\frac{(4 + 1)(\sqrt{4}+1)}{4 - 1}=\frac{5	imes(2 + 1)}{3}=\frac{15}{3}=5$. So it works. So the simplified form is $\frac{(x + 1)(\sqrt{x}+1)}{x - 1}$ or we can expand the numerator: $x\sqrt{x}+x+\sqrt{x}+1$ over $x - 1$. But maybe the problem expects us to recognize that $x + 1$ can be written as $(\sqrt{x})^2+1$, but that's not helpful. Wait, no, the correct way is to rationalize the denominator, so the simplified form is $\frac{(x + 1)(\sqrt{x}+1)}{x - 1}=\frac{x\sqrt{x}+x+\sqrt{x}+1}{x - 1}$. But maybe there's a mistake in my initial approach. Wait, another way: Let's consider that $x+1 = (\sqrt{x})^2+1$, but that's not a difference of squares. Wait, I think the correct simplified form after rationalizing is $\frac{(x + 1)(\sqrt{x}+1)}{x - 1}$.

Wait, but let's do the algebra again. Multiply numerator and denominator by $\sqrt{x}+1$:

Numerator: $(x + 1)(\sqrt{x}+1)=x\sqrt{x}+x+\sqrt{x}+1$

Denominator: $(\sqrt{x})^2-1^2=x - 1$

So the simplified form is $\frac{x\sqrt{x}+x+\sqrt{x}+1}{x - 1}$. Alternatively, we can factor the numerator as $x(\sqrt{x}+1)+1(\sqrt{x}+1)=(\sqrt{x}+1)(x + 1)$, which is what we had before. So the simplified expression is $\frac{(\sqrt{x}+1)(x + 1)}{x - 1}$ or $\frac{x\sqrt{x}+x+\sqrt{x}+1}{x - 1}$.

# Answer:
$\frac{(x + 1)(\sqrt{x}+1)}{x - 1}$ (or $\frac{x\sqrt{x}+x+\sqrt{x}+1}{x - 1}$)

### Problem 42: Simplify $\boldsymbol{\frac{x - 2}{\sqrt{x^2-4}}}$
# Explanation:
## Step1: Factor the denominator
We know that $x^2-4=(x - 2)(x + 2)$ (by the difference of squares formula $a^2-b^2=(a - b)(a + b)$), so $\sqrt{x^2-4}=\sqrt{(x - 2)(x + 2)}$. So the expression becomes:
$$\frac{x - 2}{\sqrt{(x - 2)(x + 2)}}$$
## Step2: Simplify the square root (assuming $x>2$ to ensure the radicand is positive and the denominator is non - zero)
We can rewrite the square root as $\sqrt{x - 2}\cdot\sqrt{x + 2}$ (using the property $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ for $a,b\geq0$). So the expression is:
$$\frac{x - 2}{\sqrt{x - 2}\cdot\sqrt{x + 2}}$$
## Step3: Cancel out the common factor
We can cancel out one factor of $\sqrt{x - 2}$ from the numerator and the denominator (since $x>2$, $x - 2>0$ and $\sqrt{x - 2}
eq0$). The numerator $x - 2=(\sqrt{x - 2})^2$, so:
$$\frac{(\sqrt{x - 2})^2}{\sqrt{x - 2}\cdot\sqrt{x + 2}}=\frac{\sqrt{x - 2}}{\sqrt{x + 2}}$$
## Step4: Rationalize the denominator (optional, but to simplify further)
Multiply the numerator and denominator by $\sqrt{x + 2}$:
$$\frac{\sqrt{x - 2}\cdot\sqrt{x + 2}}{\sqrt{x + 2}\cdot\sqrt{x + 2}}=\frac{\sqrt{(x - 2)(x + 2)}}{x + 2}=\frac{\sqrt{x^2-4}}{x + 2}$$
But the form $\frac{\sqrt{x - 2}}{\sqrt{x + 2}}$ is also simplified. However, the most simplified form after canceling and rationalizing (if we want a rational denominator) is $\frac{\sqrt{x^2-4}}{x + 2}$ or $\frac{\sqrt{x - 2}}{\sqrt{x + 2}}$. But let's check with $x = 3$. Original expression: $\frac{3 - 2}{\sqrt{9 - 4}}=\frac{1}{\sqrt{5}}$. The simplified form $\frac{\sqrt{3 - 2}}{\sqrt{3 + 2}}=\frac{\sqrt{1}}{\sqrt{5}}=\frac{1}{\sqrt{5}}$, and $\frac{\sqrt{9 - 4}}{3 + 2}=\frac{\sqrt{5}}{5}=\frac{1}{\sqrt{5}}$ (after rationalizing $\frac{\sqrt{5}}{5}=\frac{1}{\sqrt{5}}$). So both forms are correct. The simplest form is $\frac{\sqrt{x - 2}}{\sqrt{x + 2}}$ or $\frac{\sqrt{x^2-4}}{x + 2}$.

# Answer:
$\frac{\sqrt{x - 2}}{\sqrt{x + 2}}$ (or $\frac{\sqrt{x^2-4}}{x + 2}$)

### Problem 43: Simplify $\boldsymbol{3\sqrt{24x}-2\sqrt{54x}+\sqrt{48}}$
# Explanation:
## Step1: Simplify each square root term
- For $3\sqrt{24x}$: Factor $24x$ as $4	imes6x$, so $\sqrt{24x}=\sqrt{4	imes6x}=2\sqrt{6x}$. Then $3\sqrt{24x}=3	imes2\sqrt{6x}=6\sqrt{6x}$.
- For $-2\sqrt{54x}$: Factor $54x$ as $9	imes6x$, so $\sqrt{54x}=\sqrt{9	imes6x}=3\sqrt{6x}$. Then $-2\sqrt{54x}=-2	imes3\sqrt{6x}=-6\sqrt{6x}$.
- For $\sqrt{48}$: Factor $48$ as $16	imes3$, so $\sqrt{48}=\sqrt{16	imes3}=4\sqrt{3}$.
## Step2: Combine like terms
Now, combine the terms with $\sqrt{6x}$: $6\sqrt{6x}-6\sqrt{6x}=0$. Then we are left with the term $4\sqrt{3}$.
# Answer:
$4\sqrt{3}$

### Problem 44: Simplify $\boldsymbol{5\sqrt{18c}+3\sqrt{72c}}$ (assuming the last part of the problem is missing, but we can simplify the given terms)
# Explanation:
## Step1: Simplify each square root term
- For $5\sqrt{18c}$: Factor $18c$ as $9	imes2c$, so $\sqrt{18c}=\sqrt{9	imes2c}=3\sqrt{2c}$. Then $5\sqrt{18c}=5	imes3\sqrt{2c}=15\sqrt{2c}$.
- For $3\sqrt{72c}$: Factor $72c$ as $36	imes2c$, so $\sqrt{72c}=\sqrt{36	imes2c}=6\sqrt{2c}$. Then $3\sqrt{72c}=3	imes6\sqrt{2c}=18\sqrt{2c}$.
## Step2: Combine like terms
Add the two simplified terms: $15\sqrt{2c}+18\sqrt{2c}=(15 + 18)\sqrt{2c}=33\sqrt{2c}$.
# Answer:
$33\sqrt{2c}$ (assuming the problem is $5\sqrt{18c}+3\sqrt{72c}$)

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QUESTION IMAGE

Question

Explanation:

Problem 39: Simplify $\boldsymbol{\sqrt[3]{16y^4z^{12}}}$

Step1: Factor the radicand

Step2: Use the property of cube roots $\sqrt[3]{ab}=\sqrt[3]{a}\cdot\sqrt[3]{b}$

Step3: Simplify the perfect cube roots

Step1: Factor the radicand

Step2: Use the property of cube roots $\sqrt[3]{ab}=\sqrt[3]{a}\cdot\sqrt[3]{b}$

Step3: Simplify the perfect cube roots

Step1: Recognize the numerator as a difference of squares (or use substitution)

Step2: Simplify the denominator

Answer:

Problem 40: Simplify $\boldsymbol{\sqrt[3]{-54x^6y^{11}}}$