Question

Kết quả của phép trừ \(\dfrac{2}{{{{(x + 1)}^2}}} - \dfrac{1}{{{x^2} - 1}}\) là:

A. \(\dfrac{{3 - x}}{{(x - 1){{(x + 1)}^2}}}\)

B. \(\dfrac{{x - 3}}{{(x - 1){{(x + 1)}^2}}}\)

C. \(\dfrac{{x - 3}}{{{{(x + 1)}^2}}}\)

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

 

Answer 1

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

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

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

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

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

\(=\dfrac{x-3}{\left(x+1\right)^2\left(x-1\right)}\)

⇒Chọn B

Answer 2

\(\dfrac{2}{\left(x+1\right)^2}-\dfrac{1}{x^2-1}\\ =\dfrac{2}{\left(x+1\right)^2}-\dfrac{1}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{2.\left(x-1\right)-\left(x+1\right)}{\left(x+1\right)^2.\left(x-1\right)}\\ =\dfrac{2x-2-x-1}{\left(x+1\right)^2.\left(x-1\right)}\\ =\dfrac{x-3}{\left(x+1\right)^2\left(x-1\right)}\\ =>B\)