Ребята решите пожалуйста

Г) $\frac{4}{x+3} - \frac{5}{3-x} = \frac{1}{x-3} -1$
$\frac{4}{x+3} + \frac{5}{3+x} = -\frac{1}{x+3} -1$
$\frac{9}{3+x}= -\frac{1-x-3}{3+x}$
9=-(1-x-3)
9=-1+x+3
9=x+2
x=7
д) $\frac{3}{x}+ \frac{4}{x-1} = \frac{5-x}{ x^{2} -x} \\ \frac{3(x-1)+4x}{ x^{2} -x} =\frac{5-x}{ x^{2} -x} \\ 3x-3+4x=5-x \\ 3x+4x+x=3+5 \\ 8x=8 \\ x=1$
е) $\frac{3y-2}{y} - \frac{1}{y-2} = \frac{3y+4}{ y^{2}-y} \\ \frac{(3y-2)(y-2)-y}{y ^{2}-y } = \frac{3y+4}{ y^{2}-y} \\ 3 y^{2} -6y-2y+4-y=3y+4 \\ 3y ^{2} -12y=0 \\ 3y(y-4)=0 \\ 3y=0; y=0 \\ y-4=0;y=4$