640° Вправа 640Алгебра 8 клас Тарасенкова 2025 · § 15 · стор. 191

640. Винесіть множник з-під знака кореня:

1) ${\displaystyle y>0}$: ${\displaystyle \sqrt{7y^2}=}$${\displaystyle \left|y\right|\sqrt{7}=}$${\displaystyle y\sqrt{7}}$;

2) ${\displaystyle y<0}$: ${\displaystyle \sqrt{7y^2}=}$${\displaystyle \left|y\right|\sqrt{7}=}$${\displaystyle -y\sqrt{7}}$;

3) ${\displaystyle a<0}$, ${\displaystyle b>0}$: ${\displaystyle \sqrt{12a^2{b}^2}=}$${\displaystyle 2\left|a\right|\left|b\right|\sqrt{3}=}$${\displaystyle 2(-a)b\sqrt{3}=}$${\displaystyle -2ab\sqrt{3}}$;

4) ${\displaystyle a>0}$, ${\displaystyle b<0}$: ${\displaystyle \sqrt{12a^2{b}^2}=}$${\displaystyle 2\left|a\right|\left|b\right|\sqrt{3}=}$${\displaystyle 2a(-b)\sqrt{3}=}$${\displaystyle -2ab\sqrt{3}}$;

5) ${\displaystyle a>0}$, ${\displaystyle b>0}$: ${\displaystyle \sqrt{12a^2{b}^2}=}$${\displaystyle 2\left|a\right|\left|b\right|\sqrt{3}=}$${\displaystyle 2ab\sqrt{3}}$;

6) ${\displaystyle a<0}$, ${\displaystyle b<0}$: ${\displaystyle \sqrt{12a^2{b}^2}=}$${\displaystyle 2\left|a\right|\left|b\right|\sqrt{3}=}$${\displaystyle 2(-a)(-b)\sqrt{3}=}$${\displaystyle 2ab\sqrt{3}}$;

7) ${\displaystyle \sqrt{27m^{8}n}=}$${\displaystyle \sqrt{9\cdot 3\cdot {\left(m^4\right)}^2\cdot n}=}$${\displaystyle 3m^4\sqrt{3n}}$;

8) ${\displaystyle \sqrt{32x^4{y}^{20}}=}$${\displaystyle \sqrt{16\cdot 2\cdot {\left(x^2\right)}^2\cdot {\left(y^{10}\right)}^2}=}$${\displaystyle 4x^2{y}^{10}\sqrt{2}}$.