Question

Помогите пожалуйста алгебра только правильно а то бан​

Answer 1

$\displaystyle\bf\\\frac{x}{x+3} +\frac{x-2}{x-3} =\frac{18-4x}{x^{2} -9} \\\\\\\frac{x}{x+3} +\frac{x-2}{x-3} -\frac{18-4x}{(x -3)(x+3)} =0\\\\\\\frac{x\cdot(x-3)+(x-2)\cdot(x+3)-18+4x}{(x-3)(x+3)} =0\\\\\\\frac{x^{2} -3x+x^{2} +3x-2x-6-18+4x}{(x-3)(x+3)} =0\\\\\\\frac{2x^{2} +2x-24}{(x-3)(x+3)} =0\\\\\\\left \{ {{2x^{2} +2x-24=0} \atop {x-3\neq 0 \ , \ x+3\neq 0}} \right. \\\\\\\left \{ {{x^{2} +x-12=0} \atop {x\neq 3 \ , \ x\neq -3}} \right. \\\\\\x^{2} +x-12=0$

$\displaystyle\bf\\D=1^{2} -4\cdot(-12)=1+48=49=7^{2} \\\\\\x_{1} =\frac{-1-7}{2} =\frac{-8}{2} =-4\\\\\\x_{2} =\frac{-1+7}{2}=\frac{6}{2} =3-neyd \\\\\\Otvet: \ -4$