Решебник по алгебре 8 класс Мерзляк ФГОС Задание 181

 

\[\boxed{\text{181\ (181).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\frac{x^{2} + 14x + 49}{x + 6}\ :\]

\[:\left( \frac{13}{x + 6} - x^{\backslash x + 6} + 6^{\backslash x + 6} \right) =\]

\[= \frac{(x + 7)^{2}}{x + 6}\ :\]

\[:\frac{13 - x^{2} - 6x + 6x + 36}{x + 6} =\]

\[= \frac{(x + 7)^{2}}{x + 6}\ :\frac{49 - x^{2}}{x + 6} =\]

\[= \frac{(x + 7)^{2}}{x + 6} \cdot \frac{x + 6}{49 - x^{2}} =\]

\[= \frac{(x + 7)^{2}(x + 6)}{(x + 6)(7 + x)(7 - x)} = \frac{x + 7}{7 - x}\]

\[2)\ \left( c^{\backslash c + 8} - \frac{2c - 9}{c + 8} \right):\frac{c^{2} + 3c}{c^{2} - 64} +\]

\[+ \frac{24}{c} = \frac{c^{2} + 8c - 2c + 9}{c + 8} \cdot\]

\[\cdot \frac{c^{2} - 64}{c^{2} + 3c} + \frac{24}{c}\]

\[= \frac{24}{c} + \frac{(c + 3)^{2}(c - 8)(c + 8)}{(c + 8)c(c + 3)} =\]

\[= \frac{24}{c} + \frac{(c + 3)(c - 8)}{c} =\]

\[= \frac{24 + c^{2} + 3c - 8c - 24}{c} =\]

\[= \frac{c^{2} - 5c}{c} = \frac{c(c - 5)}{c} = c - 5\]

\[3)\ \left( \frac{36}{x^{2} - 9} - \frac{x - 3^{\backslash x - 3}}{x + 3} - \frac{3 + x^{\backslash\text{-}(x + 3)}}{3 - x} \right)\ :\]

\[:\frac{6}{3 - x} =\]

\[= \frac{36 - x^{2} + 6x - 9 + 9 + 6x + x^{2}}{x^{2} - 9} \cdot\]

\[\cdot \frac{3 - x}{6} =\]

\[= \frac{(12x + 36) \cdot (3 - x)}{(x - 3)(x + 3) \cdot 6} =\]

\[= \frac{12(x + 3)(3 - x)}{6 \cdot (x - 3)(x + 3)} = - 2\]

\[4)\ \left( \frac{2y - 1^{\backslash y - 2}}{y^{2} + 2y + 4} + \frac{9y + 6}{y^{3} - 8} + \frac{1^{\backslash y^{2} + 2y + 4}}{y - 2} \right) \cdot\]

\[\cdot \frac{y^{2} - 4}{18} =\]

\[= \frac{2y^{2} - 5y + 2 + 9y + 6 + y^{2} + 2y + 4}{y^{3} - 8} \cdot\]

\[\cdot \frac{y^{2} - 4}{18} =\]

\[= \frac{3\left( y^{2} + 2y + 4 \right)(y - 2)(y + 2)}{18(y - 2)\left( y^{2} + 2y + 4 \right)} =\]

\[= \frac{y + 2}{6}\]

\[\boxed{\text{181.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\left( \frac{b}{a^{2} - ab} - \frac{2}{a - b} - \frac{a}{b^{2} - ab} \right)\ :\]

\[:\frac{a^{2} - b^{2}}{4ab} = \frac{4}{a + b}\]

\[Упростим\ левую\ часть\ \]

\[равенства:\]

\[\left( \frac{b^{\backslash b}}{a(a - b)} - \frac{2^{\backslash ab}}{a - b} + \frac{a^{\backslash a}}{b(a - b)} \right)\ :\]

\[:\frac{a^{2} - b^{2}}{4ab} = \frac{4}{a + b}\]

\[\frac{b^{2} - 2ab + a^{2}}{\text{ab}(a - b)}\ :\frac{a^{2} - b^{2}}{4ab} = \frac{4}{a + b}\]

\[\frac{(b - a)^{2} \cdot 4ab}{ab(a - b)(a - b)(a + b)} = \frac{4}{a + b}\]

\[\frac{4}{a + b} = \frac{4}{a + b}.\]

\[Тождество\ доказано.\]

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

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

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

\[Упростим\ левую\ часть\ \]

\[равенства:\]

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

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

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

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

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

\[\frac{3(a + b)}{(a + b)} = 3\]

\[3 = 3.\]

\[Тождество\ доказано.\]