ГДЗ по алгебре 8 класс Макарычев Задание 323

\[\boxed{\text{323\ (323).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ 16 + x^{2} = 0\]

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

\[x = \varnothing.\]

\[\textbf{б)}\ 0,3x^{2} = 0,027\]

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

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

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

\[x = \pm 0,3.\]

\[\textbf{в)}\ 0,5x^{2} = 30\]

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

\[x^{2} = 60\]

\[x = \pm \sqrt{60}\]

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

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

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

\[x = \varnothing.\]

\[\textbf{д)}\ x^{3} - 3x = 0\]

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

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

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

\[\textbf{е)}\ x^{3} - 11x = 0\]

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

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

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