Question

Thực hiện phép tính và rút gọn:

a) \(\dfrac{1}{p^2-4p-5}+\dfrac{2p}{p-5}\)

b) \(x+\dfrac{3x}{x+2}+2\)

c) \(\dfrac{x}{x+y}+\dfrac{4}{x^2+3xy+2y^2}+\dfrac{-3x}{x+2y}\)

Answer 1

Lời giải:
a)

\(\frac{1}{p^2-4p-5}+\frac{2p}{p-5}=\frac{1}{p(p-5)+(p-5)}+\frac{2p}{p-5}\)

\(=\frac{1}{(p+1)(p-5)}+\frac{2p(p+1)}{(p-5)(p+1)}=\frac{1+2p(p+1)}{(p+1)(p-5)}\)

\(=\frac{2p^2+2p+1}{p^2-4p-5}\)

b) \(x+\frac{3x}{x+2}+2=(x+2)+\frac{3x}{x+2}=\frac{(x+2)^2+3x}{x+2}\)

\(=\frac{x^2+7x+4}{x+2}\)

c) \(\frac{x}{x+y}+\frac{4}{x^2+3xy+2y^2}+\frac{-3x}{x+2y}\)

\(=\frac{x(x+2y)}{(x+y)(x+2y)}+\frac{4}{(x+y)(x+2y)}+\frac{-3x(x+y)}{(x+y)(x+2y)}\)

\(=\frac{x(x+2y)+4-3x(x+y)}{(x+y)(x+2y)}\)

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