Question

решите уравнение пожалуйста

Answer 1

$\frac{3+x}{4x^{2}-9 }-\frac{3-x}{4x^{2}+12x+9 }=\frac{2}{2x-3}\\\\\frac{3+x}{(2x-3)(2x+3)}-\frac{3-x}{(2x+3)^{2}}-\frac{2}{2x-3}=0\\\\\frac{(3+x)(2x+3)-(3-x)(2x-3)-2(4x^{2}+12x+9) }{82x-3)(2x+3)^{2} }=0\\\\\frac{6x+9+2x^{2}+3x-6x+9+2x^{2}-3x-8x^{2}-24x-18}{(2x-3)(2x+3)^{2} }\\\\\frac{-4x^{2}-24x }{(2x-3)(2x+3)^{2}}=0$

$\frac{-4x(x+6)}{(2x-3)(2x+3)^{2} }=0\\\\x_{1}=0\\\\x+6=0\\\\x_{2}=-6\\\\2x-3\neq 0;2x+3\neq 0\\\\x\neq 1,5,x\neq-1,5$

Ответ :  - 6 ; 0