Question

Помогите решить №1 через ОДЗ

Answer 1

$\frac{x^2}{x^2-9} =\frac{12-x}{x^2-9} (ODZ: x\neq -3; x\neq 3)\\x^2=12-x\\x^2-12+x=0\\x^2+4x-3x-12=0\\x(x+4)-3(x+4)=0\\(x+4)(x-3)=0\\x_{1} \neq -4\\x_{2} =3$              $\frac{6}{x-2} +\frac{5}{x} =3 (ODZ: x\neq 2; x\neq 0\\\frac{6}{x-2} +\frac{5}{x} -3=0\\\frac{6x+5x-10-3x^2+6x}{x(x-2)} =0\\17x-10-3x^2=0\\-3x^2+17x-10=0\\3x^2-2x-15x+10=0\\x(3x-2)-5(3x-2)=0\\(3x-2)(x-5)=0\\x_{1} =\frac{2}{3} ; x_{2} =5$