Решебник по алгебре 8 класс Мерзляк ФГОС Задание 178

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Год:2024
Тип:учебник
Серия:Алгоритм успеха

Задание 178

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Алгебра 8 класс ФГОС Мерзляк, Полонский, Якир Вентана-Граф 2020-2021
 
фгос Мерзляк ФГОС
Издание 1
Алгебра 8 класс ФГОС Мерзляк, Полонский, Якир Вентана-Граф 2020-2021

\[\boxed{\text{178\ (178).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\frac{a + 2}{a^{2} - 2a + 1}\ :\frac{a^{2} - 4}{3a - 3} -\]

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

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

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

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

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

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

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

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

\[= \frac{- 3}{a - 1} = \frac{3}{1 - a}\ \]

\[2)\ \frac{b^{2} + 3b}{b^{3} + 9b} \cdot\]

\[\cdot \left( \frac{b - 3^{\backslash b - 3}}{b + 3} + \frac{b + 3^{\backslash b + 3}}{b - 3} \right) =\]

\[= \frac{b(b + 3)}{b\left( b^{2} + 9 \right)} \cdot\]

\[\cdot \frac{b^{2} - 6b + 9 + b^{2} + 6b + 9}{(b + 3)(b - 3)} =\]

\[= \frac{b(b + 3)\left( 2b^{2} + 18 \right)}{b\left( b^{2} + 9 \right)(b + 3)(b - 3)} =\]

\[= \frac{2(b^{2} + 9)}{(b^{2} + 9)(b - 3)} = \frac{2}{b - 3}\]

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

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

\[= \frac{9c^{2} + 6c + 1 - 9c^{2} + 6c - 1}{(3c - 1)(3c + 1)}\ :\]

\[:\frac{2c}{2(3c + 1)} =\]

\[= \frac{12c}{(3c - 1)(3c + 1)} \cdot \frac{2 \cdot (3c + 1)}{2c} =\]

\[= \frac{12c \cdot 2(3c + 1)}{(3c - 1)(3c + 1) \cdot 2c} = \frac{12}{3c - 1}\]

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

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

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

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

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

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

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

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

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

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

\[5)\ \left( \frac{a - 8}{a^{2} - 10a + 25} - \frac{a}{a^{2} - 25} \right)\ :\]

\[:\frac{a - 20}{(a - 5)^{2}} =\]

\[= \left( \frac{a - 8^{\backslash a + 5}}{(a - 5)^{2}} - \frac{a^{\backslash a - 5}}{(a - 5)(a + 5)} \right):\]

\[:\frac{a - 20}{(a - 5)^{2}} =\]

\[= \frac{a^{2} - 3a - 40 - a^{2} + 5a}{(a - 5)^{2}(a + 5)} \cdot\]

\[\cdot \frac{(a - 5)^{2}}{a - 20} =\]

\[= \frac{2(a - 20)(a - 5)^{2}}{(a - 5)^{2}(a + 5)(a - 20)} =\]

\[= \frac{2}{a + 5}\]

\[6)\left( \ \frac{2x + 1}{x^{2} + 6x + 9} - \frac{x - 2}{x^{2} + 3x} \right)\ :\]

\[:\frac{x^{2} + 6}{x^{3} - 9x} =\]

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

\[:\frac{x^{2} + 6}{x\left( x^{2} - 9 \right)} =\]

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

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

\[= \frac{\left( x^{2} + 6 \right)x(x - 3)(x + 3)}{\left( x^{2} + 6 \right)x(x + 3)^{2}} =\]

\[= \frac{x - 3}{x + 3}\]

Издание 2
фгос Мерзляк ФГОС

\[\boxed{\text{178.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\[= 2\]

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

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

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

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

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

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

\[= a - 2\]

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