Question

Cho biểu thức A=(\(\dfrac{x-2}{x^2-1}\)-\(\dfrac{x+2}{x^2+2x+1}\)).\(\dfrac{x^2-1}{2}\)

a, Rút gọn A;

b, Tính giá trị của A tại x=\(\dfrac{-1}{2}\)

Answer 1

a) \(A=\left(\dfrac{x-2}{x^2-1}-\dfrac{x+2}{x^2+2x+1}\right).\dfrac{x^2-1}{2}\left(ĐKXĐ:x\ne\pm1\right)\)

\(A=\left[\dfrac{x-2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+2}{\left(x+1\right)^2}\right].\dfrac{\left(x+1\right)\left(x-1\right)}{2}\)

\(A=\left[\dfrac{\left(x-2\right)\left(x+1\right)-\left(x+2\right)\left(x-1\right)}{\left(x+1\right)^2\left(x-1\right)}\right].\dfrac{\left(x+1\right)\left(x-1\right)}{2}\)

\(A=\left[\dfrac{x^2+x-2x-2-\left(x^2-x+2x-2\right)}{\left(x+1\right)^2\left(x-1\right)}\right].\dfrac{\left(x+1\right)\left(x-1\right)}{2}\)

\(A=\left[\dfrac{x^2-x-2-x^2-x+2}{\left(x+1\right)^2\left(x-1\right)}\right].\dfrac{\left(x+1\right)\left(x-1\right)}{2}\)

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

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

P/s: câu b bn tự làm nha

Answer 2

a.

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

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

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

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

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

\(A=\dfrac{-x}{x+1}\)

Answer 3

b. Thay \(x=\dfrac{-1}{2}\) vào A ta được:

\(\dfrac{0,5}{-0,5+1}=1\)

Vậy A = 1 khi \(x=\dfrac{-1}{2}\)