Question

Помогите пожалуйста

Answer 1

$\frac{y+3}{y-3}=\frac{2y+3}{y}\\\\y\neq3 \ ; \ y\neq0\\\\\frac{y+3}{y-3}-\frac{2y+3}{y}=0\\\\\frac{y^{2}+3y-2y^{2}-3y+6y+9}{y(y-3)}=0\\\\\frac{-y^{2} +6y+9 }{y(y-3)}=0\\\\y^{2}-6y-9=0\\\\D=(-6)^{2} -4*(-9)=36+36=72=(6\sqrt{2})^{2}\\\\y_{1}=\frac{6+6\sqrt{2}}{2}=3+3\sqrt{2}\\\\y_{2}=\frac{6-6\sqrt{2}}{2}=3-3\sqrt{2}$