Question

СРОЧНО ПОМОГИТЕ ПОЖАЛУЙСТА!
С решением ​

Answer 1

$\frac{3y+2}{4y^{2}+y}+ \frac{y-3}{16y^{2}-1 }=\frac{3}{4y-1}\\\\\frac{3y+2}{y(4y+1)}+\frac{y-3}{(4y+1)(4y-1)}-\frac{3}{4y-1}=0\\\\\frac{(3y+2)(4y-1)+(y-3)*y-3y*(4y+1)}{y(4y+1)(4y-1)}=0\\\\\frac{12y^{2}-3y+8y-2+y^{2}-3y-12y^{2}-3y}{y(4y+1)(4y-1)}=0\\\\\frac{y^{2}-y-2 }{y(4y+1)(4y-1)}=0\\\\\left \{ {{y^{2}-y-2=0 } \atop {y\neq0;y\neq-0,25;y\neq0,25}} \right.\\\\y^{2} -y-2=0\\\\y_{1}=2\\\\y_{2}=-1-teorema.Vieta\\\\Otvet:\boxed{-1;2}$