Question

алгебра за 8 класс кто нибудь​

Answer 1

$\frac{2y-2}{y+3} -\frac{18}{y^2-9} =\frac{y-6}{y-3}, y\neq -3, y\neq 3 \\\frac{2y-2}{y+3} -\frac{18}{(y-3)(y+3)}-\frac{y-6}{y-3} =0\\\frac{(2y-2)(y-3)}{(y+3)(y-3)}-\frac{18}{(y+3)(y-3)}-\frac{(y-6)(y+3)}{(y+3)(y-3)}=0 \\\frac{(2y^2-6y-2y+6)-18-(y^2+3y-6y-18)}{(y-3)(y+3)} =0 \\\frac{2y^2-6y-2y+6-18-y^2-3y+6y+18}{(y-3)(y+3)} =0\\\\\frac{y^2-5y+6}{(y-3)(y+3)} =0\\y^2-5y+6\\D=(-5)^2-4*6=1\\x1 = \frac{5-1}{2} =2\\x2=\frac{5+1}{2} =3$

Последний корень отпадает, т.к. y =/= 3.

Ответ: 2.