Question

Bài 1: Cho \(\text{A}=\dfrac{3}{2x+2}+\dfrac{5x}{x^2-1}-\dfrac{5}{2x-2}\)

a. Rút gọn

b. Tìm x để \(\dfrac{P}{2}=\dfrac{3}{x^2+2}\)

Bài 2: Chứng minh rằng (\(\left(\dfrac{x^3-y^3}{x-y}+xy\right).\left(\dfrac{x-y}{x^2-y^2}\right)^2=1\)

Answer 1

Bài 2: 

a: \(A=\dfrac{3}{2\left(x+1\right)}+\dfrac{10x}{2\left(x-1\right)\left(x+1\right)}-\dfrac{5}{2\left(x-1\right)}\)

\(=\dfrac{3x-3+10x-5x-5}{2\left(x-1\right)\left(x+1\right)}=\dfrac{8x-8}{2\left(x-1\right)\left(x+1\right)}=\dfrac{4}{x+1}\)

b: Để P/2=3/x^2+2 thì \(\dfrac{4}{2x+2}=\dfrac{3}{x^2+2}\)

\(\Leftrightarrow\dfrac{2}{x+1}=\dfrac{3}{x^2+2}\)

=>\(2x^2+4-3x-3=0\)

=>2x^2-3x+1=0

=>(x-1)(2x-1)=0

=>x=1/2(nhận) hoặc x=1(loại)