ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 316

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Пояснение.

Решение.

\[При\ переносе\ чисел\ из\ одной\ \]

\[части\ уравнения\ в\ другую\ \]

\[меняем\ знак\ \]

\[на\ противоположный.\ \]

\[\textbf{а)}\ 16 + x^{2} = 0\]

\[x^{2} = - 16\]

\[x = \varnothing.\]

\[\textbf{б)}\ 0,3x^{2} = 0,027\]

\[x^{2} = 0,027\ :0,3 = 0,09\]

\[x = \sqrt{0,09} = \pm 0,3.\]

\[\textbf{в)}\ 0,5x^{2} = 30\]

\[x^{2} = 30\ :0,5 = 60\]

\[x = \sqrt{60} = \pm 2\sqrt{15}.\]

\[\textbf{г)} - 5x^{2} = \frac{1}{20}\]

\[x^{2} = \frac{1}{20}\ :( - 5) < 0\]

\[x = \varnothing.\]

\[\textbf{д)}\ x^{3} - 3x = 0\]

\[x\left( x^{2} - 3 \right) = 0\]

\[x = 0\ \ \ \ \ \ \ x^{2} = 3\]

\[x = 0\ \ \ \ \ \ \ x = \pm \sqrt{3}.\]

\[\textbf{е)}\ x^{3} - 11x = 0\]

\[x\left( x^{2} - 11 \right) = 0\]

\[x = 0\ \ \ \ x^{2} = 11\]

\[x = 0\ \ \ \ x = \pm \sqrt{11}.\]