Решебник по алгебре 8 класс Макарычев ФГОС Задание 313

 

\[\boxed{\text{313\ (313).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[Если\ в\ результате\ \]

\[преобразования\ получается\ \]

\[равенство\ вида\ \sqrt{x} = a;\]

\[где\ a < 0;то\ такие\ равенства\ \]

\[не\ являются\ верными\ ни\ \]

\[при\ каких\ \text{x.\ }\]

\[\textbf{а)}\ \sqrt{x} = 11\]

\[x = 11^{2}\]

\[x = 121.\]

\[\textbf{б)}\ 10\sqrt{x} = 3\]

\[\sqrt{x} = \frac{3}{10}\]

\[x = \left( \frac{3}{10} \right)^{2}\]

\[x = \frac{9}{100} = 0,09.\]

\[\textbf{в)}\ \sqrt{x} = - 20\]

\[x = \varnothing.\]

\[\textbf{г)}\ 2\sqrt{x} - 1 = 0\]

\[2\sqrt{x} = 1\]

\[\sqrt{x} = \frac{1}{2}\]

\[x = \left( \frac{1}{2} \right)^{2}\]

\[x = \frac{1}{4} = 0,25.\]

\[\textbf{д)}\ 5 - \sqrt{x} = 0\]

\[\sqrt{x} = 5\]

\[x = 5^{2}\]

\[x = 25.\]

\[\textbf{е)}\ 2 + \sqrt{x} = 0\]

\[\sqrt{x} = - 2\]

\[x = \varnothing.\ \]

\[\boxed{\text{313.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

Пояснение.

Решение.

\[\textbf{а)}\ x^{2} = 36\]

\[x = \sqrt{36}\]

\[x = \pm 6.\]

\[\textbf{б)}\ x^{2} = 0,49\]

\[x = \sqrt{0,49}\]

\[x = \pm 0,7.\]

\[\textbf{в)}\ x^{2} = 121\]

\[x = \sqrt{121}\]

\[x = \pm 11.\]

\[\textbf{г)}\ x² = 11\]

\[x = \pm \sqrt{11}.\]

\[\textbf{д)}\ x^{2} = 8\]

\[x = \sqrt{8}\]

\[x = \pm 2\sqrt{2}.\]

\[\textbf{е)}\ x² = 2,5\]

\[x = \sqrt{2,5}\]

\[x = \pm 5\sqrt{0,1}.\]