Question

2y-2 / y+3 - 18 / $y^{2}$ - 9 =y-6 / y-3 помогите)))))))

Answer 1

$\frac{2y-2}{y+3}-\frac{18}{y^2-9}=\frac{y-6}{y-3}\; ;\; \; \; ODZ:\; x\ne \pm 3\\\\\frac{2(y-1)}{y+3}-\frac{18}{(y-3)(y+3)}-\frac{y-6}{y-3}=0\\\\\frac{2(y-1)(y-3)-18-(y-6)(y+3)}{(y-3)(y+3)}=0\\\\\frac{2(y^2-4y+3)-18-(y^2-3y-18)}{(y-3)(y+3)}=0\\\\\frac{y^2-5y+6}{(y-3)(y+3)}=0\\\\y^2-5y+6=0\\\\y_1=2,\; \; y_2=3\notin ODZ\\\\Otvet:\; x=2$

Answer 2

$\frac{2y-2}{y+3} - \frac{18}{y^{2} - 9} = \frac{y-6}{y-3}$

$\frac{2y-2}{y+3} - \frac{18}{(y-3)(y+3)} = \frac{y-6}{y-3}$

Приводим дроби к общему знаменателю.
$\frac{(2y-2)(y-3)}{(y+3)(y-3)} - \frac{18}{(y-3)(y+3)} - \frac{(y-6)(y+3)}{(y-3)(y+3)}=0$

$\frac{(2y-2)(y-3)-18-(y-6)(y+3)}{(y+3)(y-3)} =0 \\ \\ \frac{2y ^{2} -2y-6y+6-18-y ^{2} +6y-3y+18}{(y+3)(y-3)} =0 \\ \\ \frac{y ^{2} -5y+6}{(y+3)(y-3)} =0 \\ \\ \left \{ {{y ^{2} -5y+6=0}\atop {(y+3)(y-3) \neq 0}} \right. \left \{ {{y=2; y=3} \atop {y \neq 3; y \neq -3}} \right.$
Ответ. у=2