Question

2/у^2-3у - 1/у-3=5/у^3 -9у

Answer 1

$\frac{2}{ y^{2}-3 y} - \frac{1}{y-3} = \frac{5}{ y^{3} -9y} \\ \\ \frac{2}{y(y-3)}- \frac{1}{y-3}= \frac{5}{y(y-3)(y+3)} \\ \\ \frac{2(y+3)-y(y+3)-5}{y(y-3)(y+3)}=0 \\ \\ y \neq 0, y \neq 3, y \neq -3$

2y+6-y²-3y-5=0
-y² -y+1=0
y²+y-1=0
D=1²+4=5≈±2,24
у1=(-1-2,24)/2= -1,62
у2=(-1+2,24)/2=0,62