Question

Tính P=( \(\dfrac{a+1}{a+1-\sqrt{a^2-1}}+\dfrac{a-1}{\sqrt{a^2-1}-a+1}\)) \(\left(a-\sqrt{a^2-1}\right)\)

Answer 1

đặt \(x=\sqrt{a^2-1}\)

Ta có: \(P=\left(\dfrac{a+1}{a-x+1}+\dfrac{a-1}{a-x-1}\right)\left(a-x\right)\)

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

\(=\dfrac{-2x\left(a-x\right)}{\left(a-x\right)^2-1}=\dfrac{2x^2-2ax}{\left(x^2+a^2-1\right)-2ax}=\dfrac{2x^2-2ax}{2x^2-2ax}=1\)

Vậy P=1

Answer 2

quên, tự đặt điều kiện cho a nha, x chắc không cần đâu ^^!