Решебник по алгебре 8 класс Мерзляк ФГОС Задание 184

 

\[\boxed{\text{184\ (184).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\left( \frac{a + 3}{a^{2} - 1} - \frac{1}{a^{2} + a} \right):\frac{3a + 3}{a^{2} - a} =\]

\[= \left( \frac{a + 3^{\backslash a}}{(a - 1)(a + 1)} - \frac{1^{\backslash a - 1}}{a(a + 1)} \right):\]

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

\[= \frac{a^{2} + 3a - a + 1}{(a - 1)a(a + 1)}\ :\frac{3(a + 1)}{a(a - 1)} =\]

\[= \frac{(a + 1)^{2} \cdot a \cdot (a - 1)}{(a - 1) \cdot a(a + 1) \cdot 3(a + 1)} = \frac{1}{3}\]

\[Ответ:не\ зависит.\]

\[2)\ \left( \frac{a}{a^{2} - 49} - \frac{1}{a + 7} \right)\ :\]

\[:\frac{7a}{a^{2} + 14a + 49} - \frac{2}{a - 7} =\]

\[= \left( \frac{a}{(a - 7)(a + 7)} - \frac{1^{\backslash a - 7}}{a + 7} \right)\ :\]

\[:\frac{7a}{(a + 7)^{2}} - \frac{2}{a - 7} =\]

\[= \frac{a - a + 7}{(a - 7)(a + 7)} \cdot \frac{(a + 7)^{2}}{7a} -\]

\[- \frac{2}{a - 7} = \frac{7 \cdot (a + 7)^{2}}{7a(a - 7)(a + 7)} -\]

\[- \frac{2}{a - 7} =\]

\[= \frac{a + 7}{a(a - 7)} - \frac{2^{\backslash a}}{a - 7} =\]

\[= \frac{a + 7 - 2a}{a(a - 7)} = \frac{- a + 7}{a(a - 7)} =\]

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

\[Ответ:зависит.\]

\[\boxed{\text{184.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\frac{a^{\backslash a + 1} - \frac{a^{2}}{a + 1}}{a^{\backslash a + 1} - \frac{a}{a + 1}} =\]

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

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

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

=\(\frac{a(a + 1)}{(a + 1) \cdot a^{2}} = \frac{1}{a}\)

\[2)\ \frac{a^{\backslash a} - \frac{6a - 9}{a}}{1^{\backslash a} - \frac{3}{a}} = \frac{\frac{a^{2} - 6a + 9}{a}}{\frac{a - 3}{a}} =\]

\[= \frac{(a - 3)^{2}}{a}\ :\frac{a - 3}{a} =\]

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

\[a - 3\]

\[3)\ \frac{1}{1 - \frac{1}{1^{\backslash a} + \frac{1}{a}}} =\]

\[= 1\ :\left( 1 - \frac{1}{\frac{a + 1}{a}} \right) =\]

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

\[= 1\ :\frac{a + 1 - a}{a + 1} =\]

\[= 1\ :\frac{1}{a + 1} = a + 1\]

\[4)\ \frac{\frac{2a - b}{b} + 1^{\backslash b}}{\frac{2a + b}{b} - 1^{\backslash b}} + \frac{3^{\backslash a} - \frac{b}{a}}{\frac{3a}{b} - 1^{\backslash b}} =\]

\[= \frac{\frac{2a - b + b}{b}}{\frac{2a + b - b}{b}} + \frac{\frac{3a - b}{a}}{\frac{3a - b}{b}} =\]

\[= \frac{2a}{b}\ :\frac{2a}{b} + \frac{3a - b}{a}\ :\frac{3a - b}{b} =\]

\[= 1 + \frac{3a - b}{a} \cdot \frac{b}{3a - b} =\]

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