637° Вправа 637Алгебра 8 клас Тарасенкова 2025 · § 15 · стор. 190

637. Внесіть множник під знак кореня:

1) ${\displaystyle 2\sqrt{x}=}$${\displaystyle \sqrt{2^2\cdot x}=}$${\displaystyle \sqrt{4x}}$;

2) ${\displaystyle 5\sqrt{y}=}$${\displaystyle \sqrt{5^2\cdot y}=}$${\displaystyle \sqrt{25y}}$;

3) ${\displaystyle 6\sqrt{a}=}$${\displaystyle \sqrt{6^2\cdot a}=}$${\displaystyle \sqrt{36a}}$;

4) ${\displaystyle x>0}$: ${\displaystyle 2x\sqrt{y}=}$${\displaystyle \sqrt{{\left(2x\right)}^2\cdot y}=}$${\displaystyle \sqrt{4x^{2}y}}$;

5) ${\displaystyle x<0}$: ${\displaystyle 2x\sqrt{y}=}$${\displaystyle -\sqrt{{\left(2x\right)}^2\cdot y}=}$${\displaystyle -\sqrt{4x^{2}y}}$;

6) ${\displaystyle a<0}$: ${\displaystyle 3a\sqrt{n^3}=}$${\displaystyle -\sqrt{{\left(3a\right)}^2\cdot n^3}=}$${\displaystyle -\sqrt{9a^2{n}^3}}$;

7) ${\displaystyle a>0}$: ${\displaystyle 3a\sqrt{n^3}=}$${\displaystyle \sqrt{{\left(3a\right)}^2\cdot n^3}=}$${\displaystyle \sqrt{9a^2{n}^3}}$;

8) ${\displaystyle m<0}$: ${\displaystyle m\sqrt{m^2}=}$${\displaystyle -\sqrt{m^2\cdot m^2}=}$${\displaystyle -\sqrt{m^4}}$;

9) ${\displaystyle m^5\ge 0}$, тому ${\displaystyle m\ge 0}$: ${\displaystyle m\sqrt{m^5}=}$${\displaystyle \sqrt{m^2\cdot m^5}=}$${\displaystyle \sqrt{m^7}}$;

10) ${\displaystyle 3y^2\sqrt{x}=}$${\displaystyle \sqrt{{\left(3y^2\right)}^2\cdot x}=}$${\displaystyle \sqrt{9y^{4}x}}$;

11) ${\displaystyle 2a^4\sqrt{c}=}$${\displaystyle \sqrt{{\left(2a^4\right)}^2\cdot c}=}$${\displaystyle \sqrt{4a^{8}c}}$;

12) ${\displaystyle m>0}$: ${\displaystyle m\sqrt{mn}=}$${\displaystyle \sqrt{m^2\cdot mn}=}$${\displaystyle \sqrt{m^{3}n}}$;

13) ${\displaystyle m<0}$: ${\displaystyle m\sqrt{mn}=}$${\displaystyle -\sqrt{m^2\cdot mn}=}$${\displaystyle -\sqrt{m^{3}n}}$;

14) ${\displaystyle x<0}$, ${\displaystyle y>0}$: ${\displaystyle 2x{y}^3\sqrt{z}=}$${\displaystyle -\sqrt{{\left(2x{y}^3\right)}^2\cdot z}=}$${\displaystyle -\sqrt{4x^2{y}^{6}z}}$.