Помогите с Алгеброй прошу вас!!

1) $\displaystyle \frac{5x+6}{5-x}+ \frac{3x+16}{x-5}= \frac{5x+6}{5-x}+ \frac{3x+16}{-(5-x)}= \frac{(5x+6)-(3x+16)}{5-x}=$

$\displaystyle= \frac{5x+6-3x-16}{5-x}= \frac{2x-10}{5-x}= \frac{-(10-2x)}{5-x}= \frac{-2(5-x)}{5-x}=-2$

2) $\displaystyle \frac{16-7x}{(x-4)^2}- \frac{x-x^2}{(4-x)^2}=$
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заметим что:

$(x-4)^2=x^2-8x+16=16-8x+x^2=(4-x)^2$
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$\displaystyle= \frac{16-7x-x+x^2}{(x-4)^2}= \frac{x^2-8x+16}{(x-4)^2}= \frac{(x-4)^2}{(x-4)^2}=1$

3)

$\displaystyle \frac{2y^2-5xy}{x^2-4y^2}- \frac{x}{2y-x}- \frac{y}{x+2y}= \frac{2y^2-5xy}{(x-2y)(x+2y)}- \frac{-x}{x-2y}- \frac{y}{x+2y}=$

$\displaystyle= \frac{2y^2-5xy+x(x+2y)-y(x-2y)}{(x-2y)(x+2y)}=$

$\displaystyle \frac{2y^2-5xy+x^2+2xy-xy+2y^2}{(x-2y)(x+2y)}= \frac{x^2-4xy+4y^2}{(x-2y)(x+2y)}=$

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