Collection of 20 algebra exercises focused on simplifying square roots containing both integers and variables with exponents.

Let's solve each of these square root expressions step by step, simplifying them as much as possible. We'll use the property:

$$
\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}
$$

and also:
$$
\sqrt{a^2} = a \quad \text{(for } a \geq 0\text{)}
$$

We'll factor out perfect squares from under the radical.

---

1. $\sqrt{18}$

$$
\sqrt{18} = \sqrt{9 \cdot 2} = \sqrt{9} \cdot \sqrt{2} = 3\sqrt{2}
$$

✔ Answer: $3\sqrt{2}$

---

2. $\sqrt{125}$

$$
\sqrt{125} = \sqrt{25 \cdot 5} = \sqrt{25} \cdot \sqrt{5} = 5\sqrt{5}
$$

✔ Answer: $5\sqrt{5}$

---

3. $\sqrt{72}$

$$
\sqrt{72} = \sqrt{36 \cdot 2} = \sqrt{36} \cdot \sqrt{2} = 6\sqrt{2}
$$

✔ Answer: $6\sqrt{2}$

---

4. $\sqrt{180}$

$$
\sqrt{180} = \sqrt{36 \cdot 5} = \sqrt{36} \cdot \sqrt{5} = 6\sqrt{5}
$$

✔ Answer: $6\sqrt{5}$

---

5. $\sqrt{a^3}$

$$
\sqrt{a^3} = \sqrt{a^2 \cdot a} = \sqrt{a^2} \cdot \sqrt{a} = a\sqrt{a}
$$

✔ Answer: $a\sqrt{a}$

---

6. $\sqrt{b^7}$

$$
\sqrt{b^7} = \sqrt{b^6 \cdot b} = \sqrt{b^6} \cdot \sqrt{b} = b^3\sqrt{b}
$$

✔ Answer: $b^3\sqrt{b}$

---

7. $\sqrt{m^{11}}$

$$
\sqrt{m^{11}} = \sqrt{m^{10} \cdot m} = \sqrt{m^{10}} \cdot \sqrt{m} = m^5\sqrt{m}
$$

✔ Answer: $m^5\sqrt{m}$

---

8. $\sqrt{75x^7y^5}$

Break down constants and variables:

- $75 = 25 \cdot 3$, so $\sqrt{75} = \sqrt{25 \cdot 3} = 5\sqrt{3}$
- $x^7 = x^6 \cdot x$, so $\sqrt{x^7} = \sqrt{x^6 \cdot x} = x^3\sqrt{x}$
- $y^5 = y^4 \cdot y$, so $\sqrt{y^5} = \sqrt{y^4 \cdot y} = y^2\sqrt{y}$

Now combine:

$$
\sqrt{75x^7y^5} = 5\sqrt{3} \cdot x^3\sqrt{x} \cdot y^2\sqrt{y} = 5x^3y^2 \sqrt{3xy}
$$

✔ Answer: $5x^3y^2\sqrt{3xy}$

---

9. $\sqrt{27a^{11}b^7}$

- $27 = 9 \cdot 3$, so $\sqrt{27} = \sqrt{9 \cdot 3} = 3\sqrt{3}$
- $a^{11} = a^{10} \cdot a$, so $\sqrt{a^{11}} = a^5\sqrt{a}$
- $b^7 = b^6 \cdot b$, so $\sqrt{b^7} = b^3\sqrt{b}$

Combine:

$$
\sqrt{27a^{11}b^7} = 3\sqrt{3} \cdot a^5\sqrt{a} \cdot b^3\sqrt{b} = 3a^5b^3 \sqrt{3ab}
$$

✔ Answer: $3a^5b^3\sqrt{3ab}$

---

10. $\sqrt{32a^7b^4}$

- $32 = 16 \cdot 2$, so $\sqrt{32} = \sqrt{16 \cdot 2} = 4\sqrt{2}$
- $a^7 = a^6 \cdot a$, so $\sqrt{a^7} = a^3\sqrt{a}$
- $b^4$ is a perfect square: $\sqrt{b^4} = b^2$

Now combine:

$$
\sqrt{32a^7b^4} = 4\sqrt{2} \cdot a^3\sqrt{a} \cdot b^2 = 4a^3b^2 \sqrt{2a}
$$

✔ Answer: $4a^3b^2\sqrt{2a}$

---

11. $\sqrt{9a^8}$

- $\sqrt{9} = 3$
- $\sqrt{a^8} = a^4$

So:
$$
\sqrt{9a^8} = 3a^4
$$

✔ Answer: $3a^4$

---

12. $\sqrt{45a^7}$

- $45 = 9 \cdot 5$, so $\sqrt{45} = 3\sqrt{5}$
- $a^7 = a^6 \cdot a$, so $\sqrt{a^7} = a^3\sqrt{a}$

$$
\sqrt{45a^7} = 3\sqrt{5} \cdot a^3\sqrt{a} = 3a^3\sqrt{5a}
$$

✔ Answer: $3a^3\sqrt{5a}$

---

13. $\sqrt{36x^2y^6}$

All parts are perfect squares:

- $\sqrt{36} = 6$
- $\sqrt{x^2} = x$
- $\sqrt{y^6} = y^3$

So:
$$
\sqrt{36x^2y^6} = 6xy^3
$$

✔ Answer: $6xy^3$

---

14. $\sqrt{12x^{20}y^8}$

- $12 = 4 \cdot 3$, so $\sqrt{12} = 2\sqrt{3}$
- $x^{20}$ is a perfect square: $\sqrt{x^{20}} = x^{10}$
- $y^8$ is a perfect square: $\sqrt{y^8} = y^4$

So:
$$
\sqrt{12x^{20}y^8} = 2\sqrt{3} \cdot x^{10} \cdot y^4 = 2x^{10}y^4\sqrt{3}
$$

✔ Answer: $2x^{10}y^4\sqrt{3}$

---

15. $-\sqrt{200}$

First simplify $\sqrt{200}$:

- $200 = 100 \cdot 2$, so $\sqrt{200} = \sqrt{100 \cdot 2} = 10\sqrt{2}$

Then apply the negative sign:

$$
-\sqrt{200} = -10\sqrt{2}
$$

✔ Answer: $-10\sqrt{2}$

---

16. $\sqrt{196}$

$196 = 14^2$, so:

$$
\sqrt{196} = 14
$$

✔ Answer: $14$

---

17. $\sqrt{63x^4y}$

- $63 = 9 \cdot 7$, so $\sqrt{63} = \sqrt{9 \cdot 7} = 3\sqrt{7}$
- $x^4$ is a perfect square: $\sqrt{x^4} = x^2$
- $\sqrt{y} = \sqrt{y}$ (no simplification)

So:
$$
\sqrt{63x^4y} = 3\sqrt{7} \cdot x^2 \cdot \sqrt{y} = 3x^2\sqrt{7y}
$$

✔ Answer: $3x^2\sqrt{7y}$

---

18. $\sqrt{6x^3}$

- $\sqrt{6} = \sqrt{6}$ (no simplification)
- $x^3 = x^2 \cdot x$, so $\sqrt{x^3} = x\sqrt{x}$

So:
$$
\sqrt{6x^3} = \sqrt{6} \cdot x\sqrt{x} = x\sqrt{6x}
$$

✔ Answer: $x\sqrt{6x}$

---

19. $\sqrt{100x^5y}$

- $\sqrt{100} = 10$
- $x^5 = x^4 \cdot x$, so $\sqrt{x^5} = x^2\sqrt{x}$
- $\sqrt{y} = \sqrt{y}$

So:
$$
\sqrt{100x^5y} = 10 \cdot x^2\sqrt{x} \cdot \sqrt{y} = 10x^2\sqrt{xy}
$$

✔ Answer: $10x^2\sqrt{xy}$

---

20. $\sqrt{80x^{100}y^{49}}$

Break it down:

- $80 = 16 \cdot 5$, so $\sqrt{80} = \sqrt{16 \cdot 5} = 4\sqrt{5}$
- $x^{100}$: $\sqrt{x^{100}} = x^{50}$
- $y^{49} = y^{48} \cdot y = (y^{24})^2 \cdot y$, so $\sqrt{y^{49}} = y^{24}\sqrt{y}$

Now combine:

$$
\sqrt{80x^{100}y^{49}} = 4\sqrt{5} \cdot x^{50} \cdot y^{24}\sqrt{y} = 4x^{50}y^{24}\sqrt{5y}
$$

✔ Answer: $4x^{50}y^{24}\sqrt{5y}$

---

✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1. $\sqrt{18}$ | $3\sqrt{2}$ |
| 2. $\sqrt{125}$ | $5\sqrt{5}$ |
| 3. $\sqrt{72}$ | $6\sqrt{2}$ |
| 4. $\sqrt{180}$ | $6\sqrt{5}$ |
| 5. $\sqrt{a^3}$ | $a\sqrt{a}$ |
| 6. $\sqrt{b^7}$ | $b^3\sqrt{b}$ |
| 7. $\sqrt{m^{11}}$ | $m^5\sqrt{m}$ |
| 8. $\sqrt{75x^7y^5}$ | $5x^3y^2\sqrt{3xy}$ |
| 9. $\sqrt{27a^{11}b^7}$ | $3a^5b^3\sqrt{3ab}$ |
| 10. $\sqrt{32a^7b^4}$ | $4a^3b^2\sqrt{2a}$ |
| 11. $\sqrt{9a^8}$ | $3a^4$ |
| 12. $\sqrt{45a^7}$ | $3a^3\sqrt{5a}$ |
| 13. $\sqrt{36x^2y^6}$ | $6xy^3$ |
| 14. $\sqrt{12x^{20}y^8}$ | $2x^{10}y^4\sqrt{3}$ |
| 15. $-\sqrt{200}$ | $-10\sqrt{2}$ |
| 16. $\sqrt{196}$ | $14$ |
| 17. $\sqrt{63x^4y}$ | $3x^2\sqrt{7y}$ |
| 18. $\sqrt{6x^3}$ | $x\sqrt{6x}$ |
| 19. $\sqrt{100x^5y}$ | $10x^2\sqrt{xy}$ |
| 20. $\sqrt{80x^{100}y^{49}}$ | $4x^{50}y^{24}\sqrt{5y}$ |

Let me know if you'd like these formatted differently or need explanations for any specific one!