ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 1111

\[\boxed{\text{1111.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[\textbf{а)}\ \ \left( \frac{a + 1}{a^{2} + 1 - 2a} + \frac{1}{a - 1} \right) \cdot\]

\[\cdot \frac{a - 1}{a} - \frac{2}{a - 1} = 0\]

\[\left( \frac{a + 1}{(a - 1)^{2}} + \frac{1^{\backslash a - 1}}{a - 1} \right) \cdot \frac{a - 1}{a} -\]

\[- \frac{2}{a - 1} = 0\]

\[\frac{a + 1 + a - 1}{(a - 1)^{2}} \cdot \frac{a - 1}{a} -\]

\[- \frac{2}{a - 1} = 0\]

\[\frac{2a}{a(a - 1)} - \frac{2}{a - 1} = 0\]

\[\frac{2}{a - 1} - \frac{2}{a - 1} = 0\]

\[0 = 0.\]

\[Тождество\ доказано.\]

\[\textbf{б)}\ \left( \frac{1 + x}{x^{2} - xy} - \frac{1 - y}{y^{2} - xy} \right) \cdot\]

\[\cdot \frac{x^{2}y - y^{2}x}{x + y} = 1\]

\[\left( \frac{1 + x^{\backslash y}}{x(x - y)} + \frac{1 - y^{\backslash x}}{y(x - y)} \right) \cdot \ \]

\[\cdot \frac{x^{2}y - y^{2}x}{x + y} = 1\]

\[\frac{y + xy + x - xy}{\text{xy}(x - y)} \cdot \frac{\text{xy}(x - y)}{x + y} = 1\]

\[\frac{y + x}{x + y} = 1\]

\[1 = 1.\]

\[Тождество\ доказано.\]

\[3a \cdot \frac{a + c - c}{a^{2} - c^{2}} - \frac{3c^{2}}{a^{2} - c^{2}} = 3\]

\[\frac{3a^{2} - 3c^{2}}{a^{2} - c^{2}} = 3\]

\[\frac{3 \cdot \left( a^{2} - c^{2} \right)}{a^{2} - c^{2}} = 3\]

\[3 = 3.\]

\[Тождество\ доказано.\]