Задание во вложении Домашнее задание!СРОЧНО!

Решите задачу:

$\frac{xy+x}{x-y}: \frac{(y+1)x^2}{x+y}*\frac{x^2-xy}{x+y}= (\frac{x(y+1)}{x-y}: \frac{(y+1)x^2}{x+y})*\frac{x(x-y)}{x+y}=$

$=(\frac{x(y+1)}{x-y}* \frac{x+y}{(y+1)x^2})*\frac{x(x-y)}{x+y}= \frac{x(y+1)*(x+y)*x*(x-y)}{(x-y)(y+1)x^2(x+y)}=1$

$\frac{3x-2}{2}+\frac{1-2x}{3}-\frac{4x+3}{7}=\frac{1}{14}$

$\frac{(3x-2)*3*7}{2*3*7}+\frac{(1-2x)*2*7}{3*2*7}-\frac{(4x+3)*2*3}{7*2*3}=\frac{1*3}{14*3}$

$21(3x-2)+14(1-2x)-6(4x+3)=3$

$63x-42+14-28x-24x-18=3$

$11x=49$

$x=\frac{49}{11}$