Решебник по алгебре 8 класс Макарычев ФГОС Задание 610

 

\[\boxed{\text{610\ (610).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ 1 + \frac{1}{3 + \frac{1}{2 + \frac{1}{5 - x^{2}}}} = 1\frac{7}{24}\]

\[1 + \frac{1}{3 + \frac{1}{\frac{10 - 2x^{2} + 1}{5 - x^{2}}}} = \frac{31}{24}\]

\[1 + \frac{1}{3 + \frac{5 - x^{2}}{11 - 2x^{2}}} = \frac{31}{24}\]

\[1 + \frac{1}{\frac{33 - 6x^{2} + 5 - x^{2}}{11 - 2x^{2}}} = \frac{31}{24}\]

\[1 \cdot \left( 38 - 7x^{2} \right) + \frac{11 - 2x^{2}}{- 7x^{2} + 38} =\]

\[= \frac{31}{24}\]

\[\frac{- 7x^{2} + 38 + 11 - 2x^{2}}{- 7x^{2} + 38} = \frac{31}{24}\]

\[24 \cdot \left( - 9x^{2} + 49 \right) =\]

\[= 31 \cdot \left( - 7x^{2} + 38 \right)\]

\[- 216x^{2} + 1176 =\]

\[= - 217x^{2} + 1178\]

\[x^{2} = 2\]

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

\[Ответ:x = \pm \sqrt{2}.\]

\[\textbf{б)}\ 1 - \frac{1}{2 + \frac{1}{1 + \frac{1}{10 - x^{2}}}} = \frac{3}{5}\]

\[1 - \frac{1}{2 + \frac{1}{\frac{10 - x^{2} + 1}{10 - x^{2}}}} = \frac{3}{5}\]

\[1 - \frac{1}{2 + \frac{10 - x^{2}}{11 - x^{2}}} = \frac{3}{5}\]

\[1 - \frac{1}{\frac{22 - 2x^{2} + 10 - x^{2}}{11 - x^{2}}} = \frac{3}{5}\]

\[1 - \frac{11 - x^{2}}{32 - 3x^{2}} = \frac{3}{5}\]

\[\frac{32 - 3x^{2} - 11 + x^{2}}{32 - 3x^{2}} = \frac{3}{5}\]

\[5 \cdot \left( 21 - 2x^{2} \right) = 3 \cdot \left( 32 - 3x^{2} \right)\]

\[105 - 10x^{2} = 96 - 9x^{2}\]

\[x^{2} = 9\]

\[x = \pm 3\ \]

\[Ответ:x = \pm 3.\]

\[\boxed{\text{610.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

Пояснение.

Формулы квадрата суммы и квадрата разности:

\[a^{2} - 2ab + b^{2} = (a - b)^{2};\]

\[a^{2} + 2ab + b^{2} = (a + b)^{2}.\]

Решение.

\[x^{2} - 6x + 11 =\]

\[= \left( x^{2} - 6x + 9 \right) + 2 =\]

\[= (x - 3)^{2} + 2 > 0;\]

\[- x^{2} + 6x - 11 =\]

\[= - \left( x^{2} - 6x + 11 \right) =\]

\[= - (x - 3)^{2} - 2 < 0.\]