Решите уравнение:
\(1)\ \frac{4y+24}{5y^2-45}+\frac{y+3}{5y^2-15y}=\frac{y-3}{y^2+3y};\)
\(2)\ \frac{y+2}{8y^3+1}-\frac{1}{4y+2}=\frac{y+3}{8y^2-4y+2}.\)
Упражнение 218
Источник заимствования: Алгебра. 8 класс. Учебник для учащихся общеобразовательных организаций / А.Г. Мерзляк, В.Б. Полонский, М.С. Якир – Вентана-Граф, 2019. – 58 c. ISBN 978-5-360-07402-1
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Решение:
\(1)\ \frac{4y+24}{5y^2-45}+\frac{y+3}{5y^2-15y}=\frac{y-3}{y^2+3y};\)
\(\phantom{1)\ }\frac{4y+24}{ 5(y^2-9)}+\frac{y+3}{ 5y(y-3)}-\frac{y-3}{ y(y+3)}=0;\)
\(\phantom{1)\ }\frac{4y+24}{ 5(y-3)(y+3)}+\frac{y+3}{ 5y(y-3)}-\frac{y-3}{ y(y+3)}=0;\)
\(\phantom{1)\ }\frac{ y(4y+24)+(y+3)^2-5(y-3)^2}{ 5y(y-3)(y+3)}=0;\)
\(\phantom{1)\ }\frac{4y^2+24y+y^2+6y+9-5(y^2-6y+9)}{ 5y(y-3)(y+3)}=0;\)
\(\phantom{1)\ }\frac{4y^2+24y+y^2+6y+9-5y^2+30y-45}{ 5y(y-3)(y+3)}=0;\)
\(\phantom{1)\ }\frac{60y-36}{ 5y(y-3)(y+3)}=0;\)
\(\phantom{1)\ }\begin{cases}60y-36=0,\\[-1ex]\\5y(y-3)(y+3)\ne0;\end{cases}\)
\(\phantom{1)\ }\begin{cases}12(5y-3)=0,\\[-1ex]\\5y(y-3)(y+3)\ne0;\end{cases}\)
\(\phantom{1)\ }\begin{cases}y=0{,}6,\\[-1ex]\\y\ne0,\\[-1ex]\\y\ne3,\\[-1ex]\\y\ne{-}3.\end{cases}\)
Ответ: \(0{,}6.\)
\(\phantom{1)\ }\frac{4y+24}{ 5(y^2-9)}+\frac{y+3}{ 5y(y-3)}-\frac{y-3}{ y(y+3)}=0;\)
\(\phantom{1)\ }\frac{4y+24}{ 5(y-3)(y+3)}+\frac{y+3}{ 5y(y-3)}-\frac{y-3}{ y(y+3)}=0;\)
\(\phantom{1)\ }\frac{ y(4y+24)+(y+3)^2-5(y-3)^2}{ 5y(y-3)(y+3)}=0;\)
\(\phantom{1)\ }\frac{4y^2+24y+y^2+6y+9-5(y^2-6y+9)}{ 5y(y-3)(y+3)}=0;\)
\(\phantom{1)\ }\frac{4y^2+24y+y^2+6y+9-5y^2+30y-45}{ 5y(y-3)(y+3)}=0;\)
\(\phantom{1)\ }\frac{60y-36}{ 5y(y-3)(y+3)}=0;\)
\(\phantom{1)\ }\begin{cases}60y-36=0,\\[-1ex]\\5y(y-3)(y+3)\ne0;\end{cases}\)
\(\phantom{1)\ }\begin{cases}12(5y-3)=0,\\[-1ex]\\5y(y-3)(y+3)\ne0;\end{cases}\)
\(\phantom{1)\ }\begin{cases}y=0{,}6,\\[-1ex]\\y\ne0,\\[-1ex]\\y\ne3,\\[-1ex]\\y\ne{-}3.\end{cases}\)
Ответ: \(0{,}6.\)
\(2)\ \frac{y+2}{8y^3+1}-\frac{1}{4y+2}=\frac{y+3}{8y^2-4y+2};\)
\(\phantom{2)\ }\frac{y+2}{ (2y+1)(4y^2-2y+1)}-\frac{1}{ 2(2y+1)}-\frac{y+3}{ 2(4y^2-2y+1)}=0;\)
\(\phantom{2)\ }\frac{ 2(y+2)-(4y^2-2y+1)-(y+3)(2y+1)}{ 2(2y+1)(4y^2-2y+1)}=0;\)
\(\phantom{2)\ }\frac{2y+4-4y^2+2y-1-(2y^2+y+6y+3)}{ 2(2y+1)(4y^2-2y+1)}=0;\)
\(\phantom{2)\ }\frac{2y+4-4y^2+2y-1-2y^2-y-6y-3}{ 2(2y+1)(4y^2-2y+1)}=0;\)
\(\phantom{2)\ }\frac{{-}6y^2-3y}{ 2(2y+1)(4y^2-2y+1)}=0;\)
\(\phantom{2)\ }\begin{cases}{-}6y^2-3y=0,\\[-1ex]\\2(2y+1)(4y^2-2y+1)\ne0;\end{cases}\)
\(\phantom{2)\ }\begin{cases}{-}3y(2y+1)=0,\\[-1ex]\\2(8y^3+1)\ne0;\end{cases}\)
\(\phantom{2)\ }\begin{cases}y=0,\\[-1ex]\\y={-}0{,}5,\\[-1ex]\\y\ne{-}0{,}5.\end{cases}\)
Ответ: \(0.\)
\(\phantom{2)\ }\frac{y+2}{ (2y+1)(4y^2-2y+1)}-\frac{1}{ 2(2y+1)}-\frac{y+3}{ 2(4y^2-2y+1)}=0;\)
\(\phantom{2)\ }\frac{ 2(y+2)-(4y^2-2y+1)-(y+3)(2y+1)}{ 2(2y+1)(4y^2-2y+1)}=0;\)
\(\phantom{2)\ }\frac{2y+4-4y^2+2y-1-(2y^2+y+6y+3)}{ 2(2y+1)(4y^2-2y+1)}=0;\)
\(\phantom{2)\ }\frac{2y+4-4y^2+2y-1-2y^2-y-6y-3}{ 2(2y+1)(4y^2-2y+1)}=0;\)
\(\phantom{2)\ }\frac{{-}6y^2-3y}{ 2(2y+1)(4y^2-2y+1)}=0;\)
\(\phantom{2)\ }\begin{cases}{-}6y^2-3y=0,\\[-1ex]\\2(2y+1)(4y^2-2y+1)\ne0;\end{cases}\)
\(\phantom{2)\ }\begin{cases}{-}3y(2y+1)=0,\\[-1ex]\\2(8y^3+1)\ne0;\end{cases}\)
\(\phantom{2)\ }\begin{cases}y=0,\\[-1ex]\\y={-}0{,}5,\\[-1ex]\\y\ne{-}0{,}5.\end{cases}\)
Ответ: \(0.\)