Question

2y делённое на y-3 + 7 делённое на y+3 + 2y квадрат делённый на 9-y квадрат

Answer 1

$\frac{2y(y+3)}{y-3}+\frac{7(y-3)}{y+3}+\frac{2y^2}{9-y^2}=\frac{2y^2+6y}{(y-3)(y+3)}+\frac{7y-21}{(y+3)(y-3)}-\frac{2y^2}{y^2-9}=$
$=\frac{2y^2+6y+7y-21}{(y-3)(y+3)}-\frac{2y^2}{(y-3)(y+3)}=\frac{13y-21}{(y-3)(y+3)}$