§ 6. Упражнение 180. «Алгебра. 8 класс. Учебник для учащихся общеобразовательных организаций» АЛГЕБРА 8 ГДЗ Упражнение 180

    Упражнение 180

    Упростите выражение:
    \(1)\ \left(\frac{15}{x-7}-x-7\right)\cdot\frac{7-x}{x^2-16x+64};\)
    \(2)\ \left(a-\frac{5a-16}{a-3}\right):\left(2a-\frac{2a}{a-3}\right);\)
    \(3)\ \left(\frac{1}{a}+\frac{2}{b}+\frac{a}{b^2}\right)\cdot\frac{ab}{a^2-b^2}+\frac{2}{b-a};\)
    \(4)\ \left(\frac{a}{a-1}-\frac{a}{a+1}-\frac{a^2+1}{1-a^2}\right):\frac{a^2+a}{(a-1)^2};\)
    \(5)\ \left(\frac{x+2y}{x-2y}-\frac{x-2y}{x+2y}-\frac{16y^2}{x^2-4y^2}\right):\frac{4y}{x+2y};\)
    \(6)\ \left(\frac{3a-8}{a^2-2a+4}+\frac{1}{a+2}-\frac{4a-28}{a^3+8}\right)\cdot\frac{a^2-4}{4}.\)
    Источник заимствования: Алгебра. 8 класс. Учебник для учащихся общеобразовательных организаций / – Вентана-Граф, 2019. – 44 c. ISBN 978-5-360-07402-1
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    Решение:

    \(1)\ \left(\frac{15}{x-7}-x-7\right)\cdot\frac{7-x}{x^2-16x+64}=\frac{x+8}{x-8}\)
    \(1)\ \frac{15}{x-7}-x-7={\frac{15}{x-7}}^{\backslash1}-{\frac{x}{1}}^{\backslash{x\ -\ 7}}-{\frac{7}{1}}^{\backslash{x\ -\ 7}}=\frac{15-(x^2-7x)-(7x-49)}{x-7}=\frac{15-x^2+7x-7x+49}{x-7}=\frac{{-}x^2+64}{x-7}=\frac{ {-}(x^2-64)}{x-7}={-}\frac{x^2-64}{x-7}=\frac{x^2-64}{7-x}\)
    \(2)\ \frac{x^2-64}{7-x}\cdot\frac{7-x}{x^2-16x+64}=\frac{ (x-8)(x+8)}{7-x}\cdot\frac{7-x}{ (x-8)^2}=\frac{x+8}{x-8}\)
    \(2)\ \left(a-\frac{5a-16}{a-3}\right):\left(2a-\frac{2a}{a-3}\right)=\frac{a-4}{2a}\)
    \(1)\ a-\frac{5a-16}{a-3}={\frac{a}{1}}^{\backslash{a\ -\ 3}}-{\frac{5a-16}{a-3}}^{\backslash1}=\frac{a^2-3a-(5a-16)}{a-3}=\frac{a^2-3a-5a+16}{a-3}=\frac{a^2-8a+16}{a-3}=\frac{ (a-4)^2}{a-3}\)
    \(2)\ 2a-\frac{2a}{a-3}={\frac{2a}{1}}^{\backslash{a\ -\ 3}}-{\frac{2a}{a-3}}^{\backslash1}=\frac{2a^2-6a-2a}{a-3}=\frac{2a^2-8a}{a-3}\)
    \(3)\ \frac{ (a-4)^2}{a-3}:\frac{2a^2-8a}{a-3}=\frac{ (a-4)^2}{a-3}\cdot\frac{a-3}{ 2a(a-4)}=\frac{a-4}{2a}\)
    \(3)\ \left(\frac{1}{a}+\frac{2}{b}+\frac{a}{b^2}\right)\cdot\frac{ab}{a^2-b^2}+\frac{2}{b-a}=\frac{1}{b}\)
    \(1)\ {\frac{1}{a}}^{\backslash{b^2}}+{\frac{2}{b}}^{\backslash{ab}}+{\frac{a}{b^2}}^{\backslash{a}}=\frac{b^2+2ab+a^2}{ab^2}=\frac{ (a+b)^2}{ab^2}\)
    \(2)\ \frac{ (a+b)^2}{ab^2}\cdot\frac{ab}{a^2-b^2}=\frac{ (a+b)^2}{ab^2}\cdot\frac{ab}{ (a-b)(a+b)}=\frac{a+b}{ b(a-b)}\)
    \(3)\ \frac{a+b}{ b(a-b)}+\frac{2}{b-a}=\frac{a+b}{ b(a-b)}+\frac{2}{ {-}(a-b)}={\frac{a+b}{ b(a-b)}}^{\backslash1}-{\frac{2}{a-b}}^{\backslash{b}}=\frac{a+b-2b}{ b(a-b)}=\frac{a-b}{ b(a-b)}=\frac{1}{b}\)
    \(4)\ \left(\frac{a}{a-1}-\frac{a}{a+1}-\frac{a^2+1}{1-a^2}\right):\frac{a^2+a}{ (a-1)^2}=\frac{a-1}{a}\)
    \(1)\ \frac{a}{a-1}-\frac{a}{a+1}-\frac{a^2+1}{1-a^2}=\frac{a}{a-1}-\frac{a}{a+1}-\frac{a^2+1}{ {-}(a^2-1)}={\frac{a}{a-1}}^{\backslash{a\ +\ 1}}-{\frac{a}{a+1}}^{\backslash{a\ -\ 1}}+{\frac{a^2+1}{ (a-1)(a+1)}}^{\backslash1}=\frac{a^2+a-(a^2-a)+a^2+1}{ (a-1)(a+1)}=\frac{a^2+a-a^2+a+a^2+1}{ (a-1)(a+1)}=\frac{a^2+2a+1}{ (a-1)(a+1)}=\frac{ (a+1)^2}{ (a-1)(a+1)}=\frac{a+1}{a-1}\)
    \(2)\ \frac{a+1}{a-1}:\frac{a^2+a}{ (a-1)^2}=\frac{a+1}{a-1}\cdot\frac{ (a-1)^2}{ a(a+1)}=\frac{a-1}{a}\)
    \(5)\ \left(\frac{x+2y}{x-2y}-\frac{x-2y}{x+2y}-\frac{16y^2}{x^2-4y^2}\right):\frac{4y}{x+2y}=2\)
    \(1)\ \frac{x+2y}{x-2y}-\frac{x-2y}{x+2y}-\frac{16y^2}{x^2-4y^2}={\frac{x+2y}{x-2y}}^{\backslash{x\ +\ 2y}}-{\frac{x-2y}{x+2y}}^{\backslash{x\ -\ 2y}}-{\frac{16y^2}{ (x-2y)(x+2y)}}^{\backslash1}=\frac{ (x+2y)^2-(x-2y)^2-16y^2}{ (x-2y)(x+2y)}=\frac{x^2+4xy+4y^2-(x^2-4xy+4y^2)-16y^2}{ (x-2y)(x+2y)}=\frac{x^2+4xy+4y^2-x^2+4xy-4y^2-16y^2}{ (x-2y)(x+2y)}=\frac{8xy-16y^2}{ (x-2y)(x+2y)}=\frac{ 8y(x-2y)}{ (x-2y)(x+2y)}=\frac{8y}{x+2y}\)
    \(2)\ \frac{8y}{x+2y}:\frac{4y}{x+2y}=\frac{8y}{x+2y}\cdot\frac{x+2y}{4y}=2\)
    \(6)\ \left(\frac{3a-8}{a^2-2a+4}+\frac{1}{a+2}-\frac{4a-28}{a^3+8}\right)\cdot\frac{a^2-4}{4}=a-2\)
    \(1)\ \frac{3a-8}{a^2-2a+4}+\frac{1}{a+2}-\frac{4a-28}{a^3+8}={\frac{3a-8}{a^2-2a+4}}^{\backslash{a\ +\ 2}}+{\frac{1}{a+2}}^{\backslash{a^2\ -\ 2a\ +\ 4}}-{\frac{4a-28}{ (a+2)(a^2-2a+4)}}^{\backslash1}=\frac{3a^2+6a-8a-16+a^2-2a+4-(4a-28)}{ (a+2)(a^2-2a+4)}=\frac{3a^2+6a-8a-16+a^2-2a+4-4a+28}{ (a+2)(a^2-2a+4)}=\frac{4a^2-8a+16}{ (a+2)(a^2-2a+4)}=\frac{ 4(a^2-2a+4)}{ (a+2)(a^2-2a+4)}=\frac{4}{a+2}\)
    \(2)\ \frac{4}{a+2}\cdot\frac{a^2-4}{4}=\frac{4}{a+2}\cdot\frac{ (a-2)(a+2)}{4}=a-2\)