§ 6. Упражнение 178. «Алгебра. 8 класс. Учебник для учащихся общеобразовательных организаций» АЛГЕБРА 8 ГДЗ Упражнение 178

    Упражнение 178

    Выполните действия:
    \(1)\ \frac{a+2}{a^2-2a+1}:\frac{a^2-4}{3a-3}-\frac{3}{a-2};\)
    \(2)\ \frac{b^2+3b}{b^3+9b}\cdot\left(\frac{b-3}{b+3}+\frac{b+3}{b-3}\right);\)
    \(3)\ \left(\frac{3c+1}{3c-1}-\frac{3c-1}{3c+1}\right):\frac{2c}{6c+2};\)
    \(4)\ \left(\frac{1}{a^2-4ab+4b^2}-\frac{1}{4b^2-a^2}\right):\frac{2a}{a^2-4b^2};\)
    \(5)\ \left(\frac{a-8}{a^2-10a+25}-\frac{a}{a^2-25}\right):\frac{a-20}{(a-5)^2};\)
    \(6)\ \left(\frac{2x+1}{x^2+6x+9}-\frac{x-2}{x^2+3x}\right):\frac{x^2+6}{x^3-9x}.\)
    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:

    \(1)\ \frac{a+2}{a^2-2a+1}:\frac{a^2-4}{3a-3}-\frac{3}{a-2}=\frac{3}{1-a}\)
    \(1)\ \frac{a+2}{a^2-2a+1}:\frac{a^2-4}{3a-3}=\frac{a+2}{ (a-1)^2}\cdot\frac{ 3(a-1)}{ (a-2)(a+2)}=\frac{3}{ (a-1)(a-2)}\)
    \(2)\ {\frac{3}{ (a-1)(a-2)}}^{\backslash1}-{\frac{3}{a-2}}^{\backslash{a\ -\ 1}}=\frac{3-(3a-3)}{ (a-1)(a-2)}=\frac{3-3a+3}{ (a-1)(a-2)}=\frac{6-3a}{ (a-1)(a-2)}=\frac{ {-}3(a-2)}{ (a-1)(a-2)}={-}\frac{3}{a-1}=\frac{3}{1-a}\)
    \(2)\ \frac{b^2+3b}{b^3+9b}\cdot\left(\frac{b-3}{b+3}+\frac{b+3}{b-3}\right)=\frac{2}{b-3}\)
    \(1)\ {\frac{b-3}{b+3}}^{\backslash{b\ -\ 3}}+{\frac{b+3}{b-3}}^{\backslash{b\ +\ 3}}=\frac{ (b-3)^2+(b+3)^2}{ (b-3)(b+3)}=\frac{b^2-6b+9+b^2+6b+9}{ (b-3)(b+3)}=\frac{2b^2+18}{ (b-3)(b+3)}\)
    \(2)\ \frac{b^2+3b}{b^3+9b}\cdot\frac{2b^2+18}{ (b-3)(b+3)}=\frac{ b(b+3)}{ b(b^2+9)}\cdot\frac{ 2(b^2+9)}{ (b-3)(b+3)}=\frac{2}{b-3}\)
    \(3)\ \left(\frac{3c+1}{3c-1}-\frac{3c-1}{3c+1}\right):\frac{2c}{6c+2}=\frac{12}{3c-1}\)
    \(1)\ {\frac{3c+1}{3c-1}}^{\backslash{3c\ +\ 1}}-{\frac{3c-1}{3c+1}}^{\backslash{3c\ -\ 1}}=\frac{ (3c+1)^2-(3c-1)^2}{ (3c-1)(3c+1)}=\frac{9c^2+6c+1-(9c^2-6c+1)}{ (3c-1)(3c+1)}=\frac{9c^2+6c+1-9c^2+6c-1}{ (3c-1)(3c+1)}=\frac{12c}{ (3c-1)(3c+1)}\)
    \(2)\ \frac{12c}{ (3c-1)(3c+1)}:\frac{2c}{6c+2}=\frac{12c}{ (3c-1)(3c+1)}\cdot\frac{ 2(3c+1)}{2c}=\frac{12}{3c-1}\)
    \(4)\ \left(\frac{1}{a^2-4ab+4b^2}-\frac{1}{4b^2-a^2}\right):\frac{2a}{a^2-4b^2}=\frac{1}{a-2b}\)
    \(1)\ \frac{1}{a^2-4ab+4b^2}-\frac{1}{4b^2-a^2}=\frac{1}{ (a-2b)^2}-\frac{1}{ {-}(a^2-4b^2)}={\frac{1}{ (a-2b)^2}}^{\backslash{a\ +\ 2b}}+{\frac{1}{ (a-2b)(a+2b)}}^{\backslash{a\ -\ 2b}}=\frac{a+2b+a-2b}{ (a-2b)^2(a+2b)}=\frac{2a}{ (a-2b)^2(a+2b)}\)
    \(2)\ \frac{2a}{ (a-2b)^2(a+2b)}:\frac{2a}{a^2-4b^2}=\frac{2a}{ (a-2b)^2(a+2b)}\cdot\frac{ (a-2b)(a+2b)}{2a}=\frac{1}{a-2b}\)
    \(5)\ \left(\frac{a-8}{a^2-10a+25}-\frac{a}{a^2-25}\right):\frac{a-20}{ (a-5)^2}=\frac{2}{a+5}\)
    \(1)\ \frac{a-8}{a^2-10a+25}-\frac{a}{a^2-25}={\frac{a-8}{ (a-5)^2}}^{\backslash{a\ +\ 5}}-{\frac{a}{ (a-5)(a+5)}}^{\backslash{a\ -\ 5}}=\frac{a^2+5a-8a-40-(a^2-5a)}{ (a-5)^2(a+5)}=\frac{a^2+5a-8a-40-a^2+5a}{ (a-5)^2(a+5)}=\frac{2a-40}{ (a-5)^2(a+5)}\)
    \(2)\ \frac{2a-40}{ (a-5)^2(a+5)}:\frac{a-20}{ (a-5)^2}=\frac{ 2(a-20)}{ (a-5)^2(a+5)}\cdot\frac{ (a-5)^2}{a-20}=\frac{2}{a+5}\)
    \(6)\ \left(\frac{2x+1}{x^2+6x+9}-\frac{x-2}{x^2+3x}\right):\frac{x^2+6}{x^3-9x}=\frac{x-3}{x+3}\)
    \(1)\ \frac{2x+1}{x^2+6x+9}-\frac{x-2}{x^2+3x}={\frac{2x+1}{ (x+3)^2}}^{\backslash{x}}-{\frac{x-2}{ x(x+3)}}^{\backslash{x\ +\ 3}}=\frac{2x^2+x-(x^2+3x-2x-6)}{ x(x+3)^2}=\frac{2x^2+x-x^2-3x+2x+6}{ x(x+3)^2}=\frac{x^2+6}{ x(x+3)^2}\)
    \(2)\ \frac{x^2+6}{ x(x+3)^2}:\frac{x^2+6}{x^3-9x}=\frac{x^2+6}{ x(x+3)^2}\cdot\frac{ x(x^2-9)}{x^2+6}=\frac{x^2+6}{ x(x+3)^2}\cdot\frac{ x(x-3)(x+3)}{x^2+6}=\frac{x-3}{x+3}\)