Question

30/y^3+27 +y-5/y^2-3y+9=2/y+3

Answer 1

$\frac{30}{y^3+27}+ \frac{y-5}{y^2-3y+9}= \frac{2}{y+3} \\ \\ \frac{30}{(y^2-3y+9)(y+3)} }+ \frac{y-5}{y^2-3y+9}= \frac{2}{y+3} \\ \\ \frac{30+(y+3)(y-5)-2(y^2-3y+9)}{(y^2-3y+9)(y+3)}=0 \\ \\ -2y^2+y^2+6y+3y-5y-18+30=0 \\ \\ -y^2+4y-3=0 \\ \\ y^2-4y+3=0 \\ \\ y_{1} + y_{2}=4 \\ y_{1} * y_{2}=3 \\ \\ y_{1}=1 \\ y_{2}=3$

Ответ: y1=1, y2=3