Question

Bai 1: Cho bt

P=(\(\dfrac{x-2}{x^2-1}\)-\(\dfrac{x+2}{x^2+2x+1}\)).(\(\dfrac{1-x^2}{2}\))²

a, Rut gon P

b, Tim x de \(\dfrac{P-4}{5}\)=x

Bai 2:Chung minh bieu thuc A=\(\dfrac{3}{2}\)-\(\dfrac{x}{x^2+x+1}\) luon duong voi moi x

Answer 1

Bài 1: 

a: \(P=\left(\dfrac{x-2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+2}{\left(x+1\right)^2}\right)\cdot\dfrac{\left(x-1\right)^2\cdot\left(x+1\right)^2}{4}\)

\(=\dfrac{x^2-x-2-x^2-x+2}{\left(x-1\right)\left(x+1\right)^2}\cdot\dfrac{\left(x-1\right)^2\cdot\left(x+1\right)^2}{4}\)

\(=\dfrac{-2x}{1}\cdot\dfrac{x-1}{4}=-\dfrac{x\left(x-1\right)}{2}\)

b: Để \(\dfrac{P-4}{5}=x\) thì P-4=5x

=>P=5x+4

\(\Leftrightarrow-\dfrac{x\left(x-1\right)}{2}=5x+4\)

=>-x²+x=10x+8

=>x²-x=-10x-8

=>x²+9x+8=0

=>x=-8(nhận) hoặc x=-1(loại)