350(в), 351(г), 372(б) плиииз... срочно

$350. \\B) \frac{2}{3-a} + \frac{a^2-4}{a^2-9} * \frac{a+3}{a-2} = \frac{2}{3-a}+ \frac{(a+2)(a-2)(a+3)}{(a-3)(a+3)(a-2)} = \frac{2}{3-a}+ \frac{a+2}{a-3}=\\= \frac{2}{3-a}- \frac{a+2}{3-a}= \frac{-a}{3-a} = \frac{a}{a-3}$ $351.\\b) (y- \frac{y^2}{y+1} ): ( y- \frac{y}{y+1} )=( \frac{y^2+y-y^2}{y+1} ):( \frac{y^2+y-y}{y+1} ) = \frac{y(y+1)}{(y+1)y^2} = \frac{1}{y}$
$352.\\b) \frac{1}{y+3}+ \frac{2}{y^2+4y+3} =\frac{1}{y+3}+ \frac{2}{(y+1)(y+3)} = \frac{y+1+2}{(y+1)(y+3)}= \frac{y+3}{(y+1)(y+3)}= \frac{1}{y+1} \\y^2+4y+3=(y+1)(y+3)\\y^2+4y+3=0\\D=b^2-4ac=16- 12=4=2^2,D\ \textgreater \ 0\\y_{1,2}= \frac{-bб \sqrt{D} }{2a} = \frac{-4б2}{2} =|{ {{y=-1} \atop {y=-3}} \right.$
Ответ: $\frac{1}{y+1}$