Решебник по алгебре 9 класс Мерзляк Задание 137

 

\[\boxed{\text{137\ (137).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\frac{4x + 13^{\backslash 2}}{10} - \frac{5 + 2x^{\backslash 5}}{4} >\]

\[> \frac{6 - 7x}{20} - 2^{\backslash 20}\]

\[\frac{8x + 26 - 25 - 10x}{20} >\]

\[> \frac{6 - 7x - 40}{20}\ \ \ | \cdot 20\]

\[2x + 1 > - 7x - 34\]

\[5x > - 35\]

\[x > - 7\ \ \]

\[x \in ( - 7;\ + \infty)\]

\[x_{\min} = - 6.\]

\[2)\ (x - 1)(x + 1) -\]

\[- (x - 4)(x + 2) \geq 0\]

\[x^{2} - 1 - x^{2} - 2x + 4x + 8 \geq 0\]

\[2x \geq - 7\]

\[x \geq - 3,5\ \ \]

\[x \in \lbrack - 3,5;\ + \infty)\]

\[x_{\min} = - 3.\]

\[\boxed{\mathbf{137.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ x^{3} - 25x = 0\]

\[x\left( x^{2} - 25 \right) = 0\]

\[x(x - 5)(x + 5) = 0\]

\[x = 0;\ \ x = \pm 5.\]

\[2)\ x^{5} - 7x^{3} = 0\]

\[x^{3}\left( x^{2} - 7 \right) = 0\]

\[x^{3}\left( x - \sqrt{7} \right)\left( x + \sqrt{7} \right) = 0\]

\[x = 0;\ \ x = \pm \sqrt{7}.\]

\[3)\ 9x^{3} + x = 0\]

\[9x\left( x^{2} + 1 \right) = 0\]

\[x = 0.\]

\[4)\ x^{4} - 12x^{3} + 36x^{2} = 0\]

\[x^{2}\left( x^{2} - 12x + 36 \right) = 0\]

\[x^{2}(x - 6)^{2} = 0\]

\[x = 0;\ \ x = 6.\]