Let's work through each of the expressions one by one:

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a) $\sqrt{x^3 y^2}$

To simplify this, you need to separate the powers of $x$ and $y$ and apply the square root to each term.

For $x^3$, the square root will simplify as follows:
$\sqrt{x^3} = x^{3/2} = x^{1.5}$

For $y^2$, it becomes:
$\sqrt{y^2} = y$

Thus, the simplified form of the expression is:
$x^{1.5} y$

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c) $\sqrt{18x^6}$

We separate the constant and the variable terms:

For the constant part $18$, simplify it as:
$\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$

For $x^6$, simplify it as:
$\sqrt{x^6} = x^3$

So the expression simplifies to:
$3x^3\sqrt{2}$

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e) $\frac{\sqrt{5}}{x^4}$

This expression is already simplified as much as possible, so:
$\frac{\sqrt{5}}{x^4}$

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g) $\sqrt[3]{40x^3}$

Now, let's apply the cube root to the terms:

For $40$, it does not simplify easily, so we leave it as is.

For $x^3$, since $\sqrt[3]{x^3} = x$, we get:
$x \cdot \sqrt[3]{40}$

Thus, the expression simplifies to:
$x \cdot \sqrt[3]{40}$

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i) $\sqrt[3]{-64x^3 y^6}$

We can apply the cube root to each term separately:

For $-64$, since $\sqrt[3]{-64} = -4$, we get:
$-4$

For $x^3$, $\sqrt[3]{x^3} = x$, so we get:
$x$

For $y^6$, $\sqrt[3]{y^6} = y^2$, so we get:
$y^2$

Thus, the simplified expression is:
$-4xy^2$

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k) $\sqrt[4]{16x^4 y^8}$

We can simplify the fourth root of each part:

For $16$, $\sqrt[4]{16} = 2$, so we get:
$2$

For $x^4$, $\sqrt[4]{x^4} = x$, so we get:
$x$

For $y^8$, $\sqrt[4]{y^8} = y^2$, so we get:
$y^2$

Thus, the simplified expression is:
$2xy^2$

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m) $\sqrt[5]{-32x^4}$

We now apply the fifth root:

For $-32$, $\sqrt[5]{-32} = -2$, so we get:
$-2$

For $x^4$, $\sqrt[5]{x^4} = x^{4/5}$, so we get:
$x^{4/5}$

Thus, the simplified expression is:
$-2x^{4/5}$

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Summary of results:

• a) $x^{1.5} y$
• c) $3x^3 \sqrt{2}$
• e) $\frac{\sqrt{5}}{x^4}$
• g) $x \cdot \sqrt[3]{40}$
• i) $-4xy^2$
• k) $2xy^2$
• m) $-2x^{4/5}$

Let me know if you need further clarification or details on any of the steps!

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Further questions:

1. How do you simplify square roots with variables inside?
2. Can cube roots be simplified the same way as square roots?
3. What happens when we have negative numbers inside roots?
4. How can we handle roots with fractional powers?
5. How would you handle higher powers, such as a 6th or 7th root?

Tip: When simplifying roots, always separate constants and variables, and apply the root to each part individually.