532° Вправа 532Алгебра 8 клас Тарасенкова 2025 · § 13 · стор. 158

532. Знайдіть значення виразу:

1) ${\displaystyle \frac{\sqrt{12}}{\sqrt{3}}=}$${\displaystyle \sqrt{\frac{12}{3}}=}$${\displaystyle \sqrt{4}=}$${\displaystyle 2}$;

2) ${\displaystyle \frac{\sqrt{50}}{\sqrt{2}}=}$${\displaystyle \sqrt{\frac{50}{2}}=}$${\displaystyle \sqrt{25}=}$${\displaystyle 5}$;

3) ${\displaystyle \frac{\sqrt{125}}{\sqrt{5}}=}$${\displaystyle \sqrt{\frac{125}{5}}=}$${\displaystyle \sqrt{25}=}$${\displaystyle 5}$;

4) ${\displaystyle \frac{\sqrt{24}}{\sqrt{6}}=}$${\displaystyle \sqrt{\frac{24}{6}}=}$${\displaystyle \sqrt{4}=}$${\displaystyle 2}$;

5) ${\displaystyle \frac{\sqrt{72}}{\sqrt{2}}=}$${\displaystyle \sqrt{\frac{72}{2}}=}$${\displaystyle \sqrt{36}=}$${\displaystyle 6}$;

6) ${\displaystyle \frac{\sqrt{200}}{\sqrt{2}}=}$${\displaystyle \sqrt{\frac{200}{2}}=}$${\displaystyle \sqrt{100}=}$${\displaystyle 10}$;

7) ${\displaystyle \frac{\sqrt{44}}{\sqrt{11}}=}$${\displaystyle \sqrt{\frac{44}{11}}=}$${\displaystyle \sqrt{4}=}$${\displaystyle 2}$;

8) ${\displaystyle \frac{\sqrt{60}}{\sqrt{15}}=}$${\displaystyle \sqrt{\frac{60}{15}}=}$${\displaystyle \sqrt{4}=}$${\displaystyle 2}$.