Question

Решите плиз!!!!!!35!35!!!35!!!баллов​

Answer 1

$\frac{y}{y^{2} -9}=\frac{1}{y^{2} +3y}-\frac{3}{6y+2y^{2}}\\\\\frac{y}{(y-3)(y+3)} -\frac{1}{y(y+3)}+\frac{3}{2y(3+y)}=0\\\\\frac{2y^{2}-2(y-3)+3(y-3) }{2y(y-3)(y+3)} =0\\\\\frac{2y^{2} -2y+6+3y-9}{2y(y-3)(y+3)} =0\\\\\frac{2y^{2}+y-3}{2y(y-3)(y+3)}=0\\\\\left \{ {{2y^{2}+y-3=0 } \atop {y\neq0;y\neq-3;y\neq3}} \right.$

$2y^{2} +y-3=0\\\\D=1^{2}-4*2*(-3)=1+24=25=5^{2}\\\\y_{1}=\frac{-1-5}{4}=-1,5\\\\y_{2}=\frac{-1+5}{4}=1\\\\Otvet:-1,5;1$