ГДЗ по алгебре 8 класс Макарычев Задание 394

\[\boxed{\text{394\ (394).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \sqrt{x^{2}}\ \ при\ x = 22;\ - 35;\ \]

\[- 1\frac{2}{3};0.\]

\[\sqrt{(22)^{2}} = |22| = 22;\ \ \]

\[\sqrt{( - 35)^{2}} = | - 35| = 35;\ \ \]

\[\sqrt{\left( - 1\frac{2}{3} \right)^{2}} = \left| - 1\frac{2}{3} \right| = 1\frac{2}{3};\text{\ \ }\]

\[\sqrt{0^{2}} = |0| = 0.\]

\[\textbf{б)}\ 2\sqrt{a^{2}}\ при\ a = - 7;\ \ a = 12.\]

\[2 \cdot \sqrt{( - 7)^{2}} = 2 \cdot | - 7| = 2 \cdot 7 =\]

\[= 14;\ \ \ \]

\[2 \cdot \sqrt{12^{2}} = 2 \cdot |12| = 2 \cdot 12 = 24.\]

\[\textbf{в)}\ 0,1\sqrt{y^{2}}\ при\ y = - 15;27.\]

\[0,1 \cdot \sqrt{( - 15)^{2}} = 0,1 \cdot | - 15| =\]

\[= 0,1 \cdot 15 = 1,5;\ \ \ \ \]

\[0,1 \cdot \sqrt{27^{2}} = 0,1 \cdot |27| =\]

\[= 0,1 \cdot 27 = 2,7.\]