Решите уравнение:
\(1)\ \frac{x+5}{x^2-5x}-\frac{x-5}{2x^2+10x}=\frac{x+25}{2x^2-50};\)
\(2)\ \frac{2}{x^2-9}-\frac{1}{2x^2-12x+18}=\frac{3}{2x^2+6x};\)
\(3)\ \frac{9x+12}{x^3-64}-\frac{1}{x-4}=\frac{1}{x^2+4x+16}.\)
Упражнение 217
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:
\(1)\ \frac{x+5}{x^2-5x}-\frac{x-5}{2x^2+10x}=\frac{x+25}{2x^2-50};\)
\(\phantom{1)\ }\frac{x+5}{ x(x-5)}-\frac{x-5}{ 2x(x+5)}-\frac{x+25}{ 2(x^2-25)}=0;\)
\(\phantom{1)\ }\frac{x+5}{ x(x-5)}-\frac{x-5}{ 2x(x+5)}-\frac{x+25}{ 2(x-5)(x+5)}=0;\)
\(\phantom{1)\ }\frac{ 2(x+5)^2-(x-5)^2-x(x+25)}{ 2x(x-5)(x+5)}=0;\)
\(\phantom{1)\ }\frac{ 2(x^2+10x+25)-(x^2-10x+25)-x^2-25x}{ 2x(x-5)(x+5)}=0;\)
\(\phantom{1)\ }\frac{2x^2+20x+50-x^2+10x-25-x^2-25x}{ 2x(x-5)(x+5)}=0;\)
\(\phantom{1)\ }\frac{5x+25}{ 2x(x-5)(x+5)}=0;\)
\(\phantom{1)\ }\frac{ 5(x+5)}{ 2x(x-5)(x+5)}=0;\)
\(\phantom{1)\ }\frac{5}{ 2x(x-5)}=0.\)
Ответ: корней нет.
\(\phantom{1)\ }\frac{x+5}{ x(x-5)}-\frac{x-5}{ 2x(x+5)}-\frac{x+25}{ 2(x^2-25)}=0;\)
\(\phantom{1)\ }\frac{x+5}{ x(x-5)}-\frac{x-5}{ 2x(x+5)}-\frac{x+25}{ 2(x-5)(x+5)}=0;\)
\(\phantom{1)\ }\frac{ 2(x+5)^2-(x-5)^2-x(x+25)}{ 2x(x-5)(x+5)}=0;\)
\(\phantom{1)\ }\frac{ 2(x^2+10x+25)-(x^2-10x+25)-x^2-25x}{ 2x(x-5)(x+5)}=0;\)
\(\phantom{1)\ }\frac{2x^2+20x+50-x^2+10x-25-x^2-25x}{ 2x(x-5)(x+5)}=0;\)
\(\phantom{1)\ }\frac{5x+25}{ 2x(x-5)(x+5)}=0;\)
\(\phantom{1)\ }\frac{ 5(x+5)}{ 2x(x-5)(x+5)}=0;\)
\(\phantom{1)\ }\frac{5}{ 2x(x-5)}=0.\)
Ответ: корней нет.
\(2)\ \frac{2}{x^2-9}-\frac{1}{2x^2-12x+18}=\frac{3}{2x^2+6x};\)
\(\phantom{2)\ }\frac{2}{ (x-3)(x+3)}-\frac{1}{ 2(x^2-6x+9)}-\frac{3}{ 2x(x+3)}=0;\)
\(\phantom{2)\ }\frac{2}{ (x-3)(x+3)}-\frac{1}{ 2(x-3)^2}-\frac{3}{ 2x(x+3)}=0;\)
\(\phantom{2)\ }\frac{2\cdot2x(x-3)-x(x+3)-3(x-3)^2}{ 2x(x-3)^2(x+3)}=0;\)
\(\phantom{2)\ }\frac{4x^2-12x-x^2-3x-3(x^2-6x+9)}{ 2x(x-3)^2(x+3)}=0;\)
\(\phantom{2)\ }\frac{4x^2-12x-x^2-3x-3x^2+18x-27}{ 2x(x-3)^2(x+3)}=0;\)
\(\phantom{2)\ }\frac{3x-27}{ 2x(x-3)^2(x+3)}=0;\)
\(\phantom{2)\ }\begin{cases}3x-27=0,\\[-1ex]\\2x(x-3)^2(x+3)\ne0;\end{cases}\)
\(\phantom{2)\ }\begin{cases}x=9,\\[-1ex]\\x\ne0,\\[-1ex]\\x\ne3,\\[-1ex]\\x\ne{-}3.\end{cases}\)
Ответ: \(9.\)
\(\phantom{2)\ }\frac{2}{ (x-3)(x+3)}-\frac{1}{ 2(x^2-6x+9)}-\frac{3}{ 2x(x+3)}=0;\)
\(\phantom{2)\ }\frac{2}{ (x-3)(x+3)}-\frac{1}{ 2(x-3)^2}-\frac{3}{ 2x(x+3)}=0;\)
\(\phantom{2)\ }\frac{2\cdot2x(x-3)-x(x+3)-3(x-3)^2}{ 2x(x-3)^2(x+3)}=0;\)
\(\phantom{2)\ }\frac{4x^2-12x-x^2-3x-3(x^2-6x+9)}{ 2x(x-3)^2(x+3)}=0;\)
\(\phantom{2)\ }\frac{4x^2-12x-x^2-3x-3x^2+18x-27}{ 2x(x-3)^2(x+3)}=0;\)
\(\phantom{2)\ }\frac{3x-27}{ 2x(x-3)^2(x+3)}=0;\)
\(\phantom{2)\ }\begin{cases}3x-27=0,\\[-1ex]\\2x(x-3)^2(x+3)\ne0;\end{cases}\)
\(\phantom{2)\ }\begin{cases}x=9,\\[-1ex]\\x\ne0,\\[-1ex]\\x\ne3,\\[-1ex]\\x\ne{-}3.\end{cases}\)
Ответ: \(9.\)
\(3)\ \frac{9x+12}{x^3-64}-\frac{1}{x-4}=\frac{1}{x^2+4x+16};\)
\(\phantom{3)\ }\frac{9x+12}{ (x-4)(x^2+4x+16)}-\frac{1}{x-4}-\frac{1}{x^2+4x+16}=0;\)
\(\phantom{3)\ }\frac{9x+12-(x^2+4x+16)-(x-4)}{ (x-4)(x^2+4x+16)}=0;\)
\(\phantom{3)\ }\frac{9x+12-x^2-4x-16-x+4}{ (x-4)(x^2+4x+16)}=0;\)
\(\phantom{3)\ }\frac{4x-x^2}{ (x-4)(x^2+4x+16)}=0;\)
\(\phantom{3)\ }\begin{cases}4x-x^2=0,\\[-1ex]\\(x-4)(x^2+4x+16)\ne0;\end{cases}\)
\(\phantom{3)\ }\begin{cases}x(4-x)=0,\\[-1ex]\\x^3-64\ne0;\end{cases}\)
\(\phantom{3)\ }\begin{cases}x=0,\\[-1ex]\\x=4,\\[-1ex]\\x\ne4.\end{cases}\)
Ответ: \(0.\)
\(\phantom{3)\ }\frac{9x+12}{ (x-4)(x^2+4x+16)}-\frac{1}{x-4}-\frac{1}{x^2+4x+16}=0;\)
\(\phantom{3)\ }\frac{9x+12-(x^2+4x+16)-(x-4)}{ (x-4)(x^2+4x+16)}=0;\)
\(\phantom{3)\ }\frac{9x+12-x^2-4x-16-x+4}{ (x-4)(x^2+4x+16)}=0;\)
\(\phantom{3)\ }\frac{4x-x^2}{ (x-4)(x^2+4x+16)}=0;\)
\(\phantom{3)\ }\begin{cases}4x-x^2=0,\\[-1ex]\\(x-4)(x^2+4x+16)\ne0;\end{cases}\)
\(\phantom{3)\ }\begin{cases}x(4-x)=0,\\[-1ex]\\x^3-64\ne0;\end{cases}\)
\(\phantom{3)\ }\begin{cases}x=0,\\[-1ex]\\x=4,\\[-1ex]\\x\ne4.\end{cases}\)
Ответ: \(0.\)