§ 6. Упражнение 176. «Алгебра. 8 класс. Учебник для учащихся общеобразовательных организаций» АЛГЕБРА 8 ГДЗ Упражнение 176

    Упражнение 176

    Упростите выражение:
    \(1)\ \left(\frac{a}{3}+\frac{a}{4}\right)\cdot\frac{6}{a^2};\)
    \(2)\ \frac{a^2b}{a-b}\cdot\left(\frac{1}{b}-\frac{1}{a}\right);\)
    \(3)\ \left(1+\frac{\vphantom{0}a}{b}\right):\left(1-\frac{a}{b}\right);\)
    \(4)\ \left(\frac{a^2}{b^2}-\frac{2a}{b}+1\right)\cdot\frac{b}{a-b};\)
    \(\vphantom{\left(\frac{0}{0}\right)}5)\ \frac{a^2-ab}{b^2-1}\cdot\frac{b+1}{a}-\frac{a}{b-1};\)
    \(6)\ \left(\frac{5}{m-n}-\frac{4}{m+n}\right):\frac{m+9n}{m+n};\)
    \(7)\ \frac{x-2}{x+2}\cdot\left(x-\frac{x^2}{x-2}\right);\)
    \(\vphantom{\left(\frac{0}{0}\right)}8)\ \frac{x^2+x}{4}:\frac{x^2}{4}+\frac{x-1}{x};\)
    \(9)\ \frac{6c^2}{c^2-1}:\left(\frac{1}{c-1}+1\right);\)
    \(10)\ \left(\frac{x}{x+y}+\frac{y}{x-y}\right)\cdot\frac{x^2+xy}{x^2+y^2}.\)
    Источник заимствования: Алгебра. 8 класс. Учебник для учащихся общеобразовательных организаций / – Вентана-Граф, 2019. – 43 c. ISBN 978-5-360-07402-1
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    Решение:

    \(1)\ \left(\frac{a}{3}+\frac{a}{4}\right)\cdot\frac{6}{a^2}=\frac{7}{2a}\)
    \(1)\ {\frac{a}{3}}^{\backslash4}+{\frac{a}{4}}^{\backslash3}=\frac{4a+3a}{12}=\frac{7a}{12}\)
    \(2)\ \frac{7a}{12}\cdot\frac{6}{a^2}=\frac{7}{2a}\)
    \(2)\ \frac{a^2b}{a-b}\cdot\left(\frac{1}{b}-\frac{1}{a}\right)=a\)
    \(1)\ {\frac{1}{b}}^{\backslash{a}}-{\frac{1}{a}}^{\backslash{b}}=\frac{a-b}{ab}\)
    \(2)\ \frac{a^2b}{a-b}\cdot\frac{a-b}{ab}=a\)
    \(3)\ \left(1+\frac{a}{b}\right):\left(1-\frac{a}{b}\right)=\frac{b+a}{b-a}\)
    \(1)\ 1+\frac{a}{b}={\frac{1}{1}}^{\backslash{b}}+{\frac{a}{b}}^{\backslash1}=\frac{b+a}{b}\)
    \(2)\ 1-\frac{a}{b}={\frac{1}{1}}^{\backslash{b}}-{\frac{a}{b}}^{\backslash1}=\frac{b-a}{b}\)
    \(3)\ \frac{b+a}{b}:\frac{b-a}{b}=\frac{b+a}{b}\cdot\frac{b}{b-a}=\frac{b+a}{b-a}\)
    \(4)\ \left(\frac{a^2}{b^2}-\frac{2a}{b}+1\right)\cdot\frac{b}{a-b}=\frac{a-b}{b}\)
    \(1)\ \frac{a^2}{b^2}-\frac{2a}{b}+1={\frac{a^2}{b^2}}^{\backslash1}-{\frac{2a}{b}}^{\backslash{b}}+{\frac{1}{1}}^{\backslash{b^2}}=\frac{a^2-2ab+b^2}{b^2}\)
    \(2)\ \frac{a^2-2ab+b^2}{b^2}\cdot\frac{b}{a-b}=\frac{ (a-b)^2}{b^2}\cdot\frac{b}{a-b}=\frac{a-b}{b}\)
    \(5)\ \frac{a^2-ab}{b^2-1}\cdot\frac{b+1}{a}-\frac{a}{b-1}=\frac{b}{1-b}\)
    \(1)\ \frac{a^2-ab}{b^2-1}\cdot\frac{b+1}{a}=\frac{ a(a-b)}{ (b-1)(b+1)}\cdot\frac{b+1}{a}=\frac{a-b}{b-1}\)
    \(2)\ {\frac{a-b}{b-1}}^{\backslash1}-{\frac{a}{b-1}}^{\backslash1}=\frac{a-b-a}{b-1}=\frac{{-}b}{b-1}={-}\frac{b}{b-1}=\frac{b}{1-b}\)
    \(6)\ \left(\frac{5}{m-n}-\frac{4}{m+n}\right):\frac{m+9n}{m+n}=\frac{1}{m-n}\)
    \(1)\ {\frac{5}{m-n}}^{\backslash{m\ +\ n}}-{\frac{4}{m+n}}^{\backslash{m\ -\ n}}=\frac{5m+5n-(4m-4n)}{ (m-n)(m+n)}=\frac{5m+5n-4m+4n}{ (m-n)(m+n)}=\frac{m+9n}{ (m-n)(m+n)}\)
    \(2)\ \frac{m+9n}{ (m-n)(m+n)}:\frac{m+9n}{m+n}=\frac{m+9n}{ (m-n)(m+n)}\cdot\frac{m+n}{m+9n}=\frac{1}{m-n}\)
    \(7)\ \frac{x-2}{x+2}\cdot\left(x-\frac{x^2}{x-2}\right)={-}\frac{2x}{x+2}\)
    \(1)\ x-\frac{x^2}{x-2}={\frac{x}{1}}^{\backslash{x\ -\ 2}}-{\frac{x^2}{x-2}}^{\backslash1}=\frac{x^2-2x-x^2}{x-2}=\frac{{-}2x}{x-2}={-}\frac{2x}{x-2}\)
    \(2)\ \frac{x-2}{x+2}\cdot\left({-}\frac{2x}{x-2}\right)={-}\frac{2x}{x+2}\)
    \(8)\ \frac{x^2+x}{4}:\frac{x^2}{4}+\frac{x-1}{x}=2\)
    \(1)\ \frac{x^2+x}{4}:\frac{x^2}{4}=\frac{ x(x+1)}{4}\cdot\frac{4}{x^2}=\frac{x+1}{x}\)
    \(2)\ {\frac{x+1}{x}}^{\backslash1}+{\frac{x-1}{x}}^{\backslash1}=\frac{x+1+x-1}{x}=\frac{2x}{x}=2\)
    \(9)\ \frac{6c^2}{c^2-1}:\left(\frac{1}{c-1}+1\right)=\frac{6c}{c+1}\)
    \(1)\ \frac{1}{c-1}+1={\frac{1}{c-1}}^{\backslash1}+{\frac{1}{1}}^{\backslash{c\ -\ 1}}=\frac{1+c-1}{c-1}=\frac{c}{c-1}\)
    \(2)\ \frac{6c^2}{c^2-1}:\frac{c}{c-1}=\frac{6c^2}{ (c-1)(c+1)}\cdot\frac{c-1}{c}=\frac{6c}{c+1}\)
    \(10)\ \left(\frac{x}{x+y}+\frac{y}{x-y}\right)\cdot\frac{x^2+xy}{x^2+y^2}=\frac{x}{x-y}\)
    \(1)\ {\frac{x}{x+y}}^{\backslash{x\ -\ y}}+{\frac{y}{x-y}}^{\backslash{x\ +\ y}}=\frac{x^2-xy+xy+y^2}{ (x-y)(x+y)}=\frac{x^2+y^2}{ (x-y)(x+y)}\)
    \(2)\ \frac{x^2+y^2}{ (x-y)(x+y)}\cdot\frac{x^2+xy}{x^2+y^2}=\frac{x^2+y^2}{ (x-y)(x+y)}\cdot\frac{ x(x+y)}{x^2+y^2}=\frac{x}{x-y}\)