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