Question

$\frac{x^2+2x+2}{x+1}+\frac{x^2+8x+20}{x+4} =\frac{x^2+4x+6}{x+2} +\frac{x^2+6x+12}{x+3}$
Решить уравнение.

Answer 1

Ответ:

{-2.5;0}

Объяснение:

ODZ: x≠-1;x≠-2;x≠-3;x≠-4

$\displaystyle\frac{x^2+2x+2}{x+1}+\frac{x^2+8x+20}{x+4} =\frac{x^2+4x+6}{x+2} +\frac{x^2+6x+12}{x+3}\\\\\\\frac{x^2+2x+1+1}{x+1}+\frac{x^2+8x+16+4}{x+4} =\frac{x^2+4x+4+2}{x+2} +\frac{x^2+6x+9+3}{x+3}\\\\\\\frac{(x+1)^2+1}{x+1}+\frac{(x+4)^2+4}{x+4} =\frac{(x+2)^2+2}{x+2} +\frac{(x+3)^2+3}{x+3}\\\\\\x+1+\frac{1}{x+1} +x+4+\frac{4}{x+4}=x+2+\frac{2}{x+2}+x+3+\frac{3}{x+3}\\\\\\\frac{1}{x+1} +\frac{4}{x+4}=\frac{2}{x+2}+\frac{3}{x+3}\\\\\\\frac{1}{x+1}-\frac{2}{x+2}=\frac{3}{x+3}-\frac{4}{x+4}$

$\displaystyle\\\frac{x+2-2x-2}{(x+1)(x+2)}=\frac{3x+12-4x-12}{(x+3)(x+4)}\\\\\\\frac{-x}{(x+1)(x+2)}-\frac{-x}{(x+3)(x+4)}=0\\\\\\-x(\frac{1}{(x+1)(x+2)}-\frac{1}{(x+3)(x+4)})=0\\\\\\ x*\frac{(x+3)(x+4)-(x+1)(x+2)}{(x+1)(x+2)(x+3)(x+4)}} =0\\\\\\x_1=0\\\\{(x+3)(x+4)-(x+1)(x+2)=0$

$\displaystyle\\x^2+7x+12-x^2-3x-2=0\\\\4x+10=0\\\\x_2=-2.5$