Решебник по алгебре 8 класс Макарычев ФГОС Задание 323

 

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

Пояснение.

Решение.

\[\textbf{а)}\ 16 + x^{2} = 0\]

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

\[x = \varnothing.\]

\[\textbf{б)}\ 0,3x^{2} = 0,027\]

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

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

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

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

\[\textbf{в)}\ 0,5x^{2} = 30\]

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

\[x^{2} = 60\]

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

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

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

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

\[x = \varnothing.\]

\[\textbf{д)}\ x^{3} - 3x = 0\]

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

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

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

\[\textbf{е)}\ x^{3} - 11x = 0\]

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

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

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

\[\boxed{\text{323.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

Пояснение.

Свойство степеней:

\[\left( \mathbf{\text{ab}} \right)^{\mathbf{m}}\mathbf{=}\mathbf{a}^{\mathbf{m}}\mathbf{b}^{\mathbf{n}}\mathbf{.}\]

Решение.

\[\textbf{а)}\ \ 0,49 + 2 \cdot \left( \sqrt{0,4} \right)^{2} =\]

\[= 0,49 + 2 \cdot 0,4 = 0,49 + 0,8 =\]

\[= 1,29\]

\[\textbf{б)}\ \left( 3\sqrt{11} \right)^{2} - \sqrt{6400} =\]

\[= 9 \cdot 11 - 80 = 99 - 80 = 19\]

\[\textbf{в)}\ \left( 2\sqrt{6} \right)^{2} + \left( - 3\sqrt{2} \right)^{2} =\]

\[= 4 \cdot 6 + 9 \cdot 2 = 24 + 18 = 42\]

\[\textbf{г)} - 0,1 \cdot \left( \sqrt{120} \right)^{2} - \left( \frac{1}{2}\sqrt{20} \right)^{2} =\]

\[= - 0,1 \cdot 120 - \frac{1}{4} \cdot 20 =\]

\[= - 12 - 5 = - 17\]