Bài 1:
a)= $\frac{3x+5}{-x(5+x)}$ + $\frac{25-x}{5(5-x)}$
= $\frac{-5(3x+5)+x(25-x)}{5x(5-x)}$
= $\frac{-15x - 25 +25x - x^2}{5x(5-x)}$
= $\frac{10x - 25 - x^2}{5x(5-x)}$
= $\frac{-(x^2-10x+25)}{5x(5-x)}$
= $\frac{-x(x-5)^2}{5x[-(x-5)]}$
= $\frac{x-5}{5x}$
b) = 0 (khử các số dối bằng nhau)
c) = $\frac{x}{x^2 - 1}$ - $\frac{x}{x^2-1}$
= 0
d) = $\frac{x+2}{x-1}$ - 2.($\frac{x-9}{1-x}$)
= $\frac{x+2}{x-1}$ + $\frac{2(x-9)}{x-1}$
= $\frac{x+2}{x-1}$ + $\frac{2x-18}{x-1}$
= $\frac{3x-16}{x-1}$
Bài 2:
a) = $\frac{x^2+2}{(x-1)(x^2+x+1)}$ + $\frac{2}{x^2+x+1}$ - $\frac{1}{x-1}$
= $\frac{x^2+2+2(x-1)-(x^2+x+1)}{(x-1)(x^2+x+1)}$
= $\frac{x^2+2+2x-2-x^2-x-1}{(x-1)(x^2+x+1)}$
= $\frac{x-1}{(x-1)(x^2+x+1)}$
= $\frac{1}{x^2+x+1}$
b) = $\frac{4y.5}{4y.6x^2y}$ + $\frac{2x.7}{2x.12xy^2}$ + $\frac{3xy.11}{3xy.8xy}$ (nhân thêm vào)
= $\frac{20y}{24x^2y^2}$ + $\frac{14x}{24x^2y^2}$ + $\frac{33x}{24x^2y^2}$
= $\frac{20y + 14x + 33xy}{24x^2y^2}$
c) = $\frac{1-2x}{2x}$ + $\frac{2x}{2x-1}$ + $\frac{1}{2x(1-2x)}$
= $\frac{1-2x}{2x}$ + $\frac{2x}{2x-1}$ - $\frac{1}{2x(2x-1)}$
= $\frac{(2x-1).(1-2x)+4x^2-1}{2x(2x-1)}$
= $\frac{(2x-1).[-(2x-1)]+(2x+1).(2x-1)}{2x(2x-1)}$
= $\frac{-(2x-1).[2x-1-(2x+1)]}{2x(2x-1)}$
= $\frac{-(2x-1-2x-1)}{2x}$
= $\frac{2}{2x}$
= $\frac{1}{x}$
d) = $\frac{x}{x-2y}$ + $\frac{x}{x+2y}$ + $\frac{4xy}{(2y-x).(2y+x)}$
= $\frac{x}{x-2y}$ + $\frac{x}{x+2y}$ - $\frac{4xy}{(x-2y).(2y+x)}$
= $\frac{x(x+2y)+x(x-2y)-4xy}{(x-2y)(x+2y)}$
= $\frac{x^2+2xy+x^2-2xy-4xy}{(x-2y)(x+2y)}$
= $\frac{2x^2-4xy}{(x-2y)(x+2y)}$
= $\frac{2x(x-2y)}{(x-2y)(x+2y)}$
= $\frac{2x}{x+2y}$
