Bài 4:
a) \(=\dfrac{x-2-2x+1}{x+1}=\dfrac{-x-1}{x+1}=-1\)
b) \(=\dfrac{x-5-3-2x}{3}=\dfrac{-x-8}{3}=-\dfrac{x+8}{3}\)
c) \(=\dfrac{5.5+9b-25}{15b}=\dfrac{9b}{15b}=\dfrac{3}{5}\)
d) \(=\dfrac{x}{-y\left(x-2y\right)}+\dfrac{4y}{x\left(x-2y\right)}=\dfrac{x^2-4y^2}{-xy\left(x-2y\right)}\)
\(=\dfrac{\left(x-2y\right)\left(x+2y\right)}{-xy\left(x-2y\right)}=-\dfrac{x+2y}{xy}\)
\(4,\\ a,=\dfrac{-x-1}{x+1}=-1\\ b,=\dfrac{-x-8}{3}\\ c,=\dfrac{25+9b-25}{15b}=\dfrac{9b}{15b}=\dfrac{3}{5}\\ d,=\dfrac{x^3-2x^2y+8y^3-4xy^2}{xy\left(2y-x\right)\left(x-2y\right)}\\ =\dfrac{\left(x+2y\right)\left(x^2-2xy+4y^2\right)-2xy\left(x+2y\right)}{xy\left(2y-x\right)\left(x-2y\right)}\\ =\dfrac{\left(x+2y\right)\left(2y-x\right)^2}{xy\left(2y-x\right)\left(x-2y\right)}=\dfrac{2y-x}{xy}\)
