Решебник по алгебре 8 класс Мерзляк Задание 192

\[\boxed{\text{192\ (192).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\frac{x - a}{a - b}\text{\ \ }\]

\[если\ x = \frac{\text{ab}}{a + b}:\]

\[\frac{\frac{\text{ab}}{a + b} - a^{\backslash a + b}}{\frac{\text{ab}}{a + b} - b^{\backslash a + b}} = \frac{\frac{ab - a(a + b)}{a + b}}{\frac{ab - b(a + b)}{a + b}} =\]

\[= \frac{\frac{ab - a^{2} - ab}{a + b}}{\frac{ab - ab - b^{2}}{a + b}} = \frac{\frac{- a^{2}}{a + b}}{\frac{- b^{2}}{a + b}} =\]

\[= \frac{- a^{2} \cdot (a + b)}{(a + b) \cdot \left( - b^{2} \right)} = \frac{a^{2}}{b^{2}}.\]

\[2)\ \frac{a - bx}{b + ax}\ \]

\[если\ x = \frac{a - b}{a + b}:\]

\[\frac{a^{\backslash a + b} - \frac{ab - b^{2}}{a + b}}{b^{\backslash a + b} + \frac{a^{2} - ab}{a + b}} =\]

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

\[= \frac{\frac{a + ab - ab + b^{2}}{a + b}}{\frac{ab + b^{2} + a^{2} - ab}{a + b}} =\]

\[= \frac{\frac{a^{2} + b^{2}}{a + b}}{\frac{b^{2} + a^{2}}{a + b}} = 1.\]