Решебник по алгебре 8 класс Мерзляк ФГОС Задание 180

 

\[\boxed{\text{180\ (180).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\left( \frac{15}{x - 7} - x^{\backslash x - 7} - 7^{\backslash x - 7} \right) \cdot\]

\[\cdot \frac{7 - x}{x^{2} - 16x + 64} =\]

\[= \frac{15 - x^{2} + 7x - 7x + 49}{x - 7} \cdot\]

\[\cdot \frac{7 - x}{x^{2} - 16x + 64} = \frac{64 - x^{2}}{x - 7} \cdot\]

\[\cdot \frac{7 - x}{(x - 8)^{2}} =\]

\[= \frac{(8 - x)(8 + x)(7 - x)}{(x - 7)(x - 8)^{2}} =\]

\[= \frac{8 + x}{x - 8}\]

\[2)\left( a^{\backslash a - 3} - \ \frac{5a - 16}{a - 3} \right)\ :\]

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

\[= \frac{a^{2} - 8a + 16}{a - 3}\ :\frac{2a^{2} - 8a}{a - 3} =\]

\[= \frac{(a - 4)^{2}}{a - 3} \cdot \frac{a - 3}{2a(a - 4)} =\]

\[= \frac{(a - 4)^{2} \cdot (a - 3)}{(a - 3) \cdot 2a \cdot (a - 4)} = \frac{a - 4}{2a}\]

\[3)\ \left( \frac{1^{\backslash b^{2}}}{a} + \frac{2^{\backslash ab}}{b} + \frac{a^{\backslash a}}{b^{2}} \right) \cdot\]

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

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

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

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

\[4)\ \left( \frac{a^{\backslash a + 1}}{a - 1} - \frac{a^{\backslash a - 1}}{a + 1} - \frac{a^{2} + 1}{1 - a^{2}} \right)\ :\]

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

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

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

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

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

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

\[5)\ \left( \frac{x + 2y^{\backslash x + 2y}}{x - 2y} - \frac{x - 2y^{\backslash x - 2y}}{x + 2y} - \frac{16y^{2}}{x^{2} - 4y^{2}} \right)\ :\]

\[:\frac{4y}{x + 2y} =\]

\[= \frac{(x + 2y)^{2} - (x - 2y)^{2} - 16y^{2}}{(x - 2y)(x + 2y)}\ :\]

\[:\frac{4y}{x + 2y} =\]

\[= \frac{x^{2} + 4xy + 4y^{2} - x^{2} + 4xy - 4y^{2} - 16y^{2}}{(x - 2y)(x + 2y)}\ :\]

\[:\frac{4y}{x + 2y} =\]

\[= \frac{8xy - 16y^{2}}{(x - 2y)(x + 2y)} \cdot\]

\[\cdot \frac{x + 2y}{4y} = \frac{8y(x - 2y)(x + 2y)}{4y(x - 2y)(x + 2y)} =\]

\[= 2\]

\[6)\left( \ \frac{3a - 8^{\backslash a + 2}}{a^{2} - 2a + 4} + \frac{1^{\backslash a^{2} - 2a + 4}}{a + 2} - \frac{4a - 28}{a^{3} + 8} \right) \cdot\]

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

\[= \frac{(3a - 8)(a + 2) + a^{2} - 2a + 4 - 4a + 28}{a^{3} + 8} \cdot\]

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

\[= \frac{4a^{2} - 8a + 16}{a^{3} + 8} \cdot \frac{a^{2} - 4}{4} =\]

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

\[= a - 2\]

\[\boxed{\text{180.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

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

\[Преобразуем\ левую\ часть\ равенства:\]

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

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

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

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

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

\[Тождество\ доказано.\]

\[2)\ \left( \frac{8a}{4 - a^{2}} - \frac{a - 2}{a + 2} \right)\ :\frac{a + 2}{a} +\]

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

\[Упростим\ левую\ часть\ равенства:\]

\[\left( \frac{8}{(2 - a)(2 + a)} - \frac{a - 2^{\backslash 2 - a}}{a + 2} \right)\ :\]

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

\[\frac{8a + a^{2} - 4a + 4}{4 - a^{2}}\ :\frac{a + 2}{a} +\]

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

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

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

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

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

\[- 1 = - 1.\]

\[Тождество\ доказано.\]

\[3)\ \left( \frac{3}{36 - c^{2}} + \frac{1}{c^{2} - 12c + 36} \right) \cdot\]

\[\cdot \frac{(c - 6)^{2}}{2} + \frac{3c}{c + 6} = 2\]

\[\left( \frac{3^{\backslash 6 - c}}{(6 - c)(6 + c)} + \frac{1^{\backslash 6 + c}}{(6 - c)^{2}} \right) \cdot\]

\[\cdot \frac{(c - 6)^{2}}{2} + \frac{3c}{c + 6} = 2\]

\[Упростим\ левую\ часть\ \]

\[равенства:\]

\[\frac{18 - 3c + c + 6}{(c - 6)^{2}(c + 6)} \cdot \frac{(c - 6)^{2}}{2} +\]

\[+ \frac{3c}{c + 6} = 2\]

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

\[\frac{12 - c}{c + 6} + \frac{3c}{c + 6} = 2\]

\[\frac{12 - c + 3c}{c + 6} = 2\]

\[\frac{2 \cdot (6 + c)}{(c + 6)} = 2\]

\[2 = 2.\]

\[Тождество\ доказано.\]