659 Вправа 659Алгебра 8 клас Тарасенкова 2025 · § 15 · стор. 194

659. Внесіть множник під знак кореня:

1) ${\displaystyle -x\ge 0}$, тому ${\displaystyle x\le 0}$: ${\displaystyle x\sqrt{-x}=}$${\displaystyle -\sqrt{x^2\cdot (-x)}=}$${\displaystyle -\sqrt{-x^3}}$;

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

3) ${\displaystyle a<0}$; ${\displaystyle a^2{b}^3\ge 0}$, тому ${\displaystyle b\ge 0}$ і ${\displaystyle a{b}^3\le 0}$: ${\displaystyle a{b}^3\sqrt{a^2{b}^3}=}$${\displaystyle -\sqrt{a^2{b}^6\cdot a^2{b}^3}=}$${\displaystyle -\sqrt{a^4{b}^9}}$;

4) ${\displaystyle m^2{n}^2\sqrt{nm}=}$${\displaystyle \sqrt{m^4{n}^4\cdot nm}=}$${\displaystyle \sqrt{m^5{n}^5}}$;

5) ${\displaystyle m\ge 0}$, ${\displaystyle n\le 0}$: ${\displaystyle n\sqrt{m}+m\sqrt{-n}=}$${\displaystyle -\sqrt{n^{2}m}+\sqrt{m^2\cdot (-n)}=}$${\displaystyle -\sqrt{m{n}^2}+\sqrt{-m^{2}n}}$;

6) ${\displaystyle p\ge 0}$: ${\displaystyle 3p\sqrt{m^2}+4\sqrt{p}=}$${\displaystyle \sqrt{9p^2\cdot m^2}+\sqrt{16p}=}$${\displaystyle \sqrt{9p^2{m}^2}+\sqrt{16p}}$.