Question

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

a, rút gọn A

B,tính giá trị biểu thức khi x=1/2

Answer 1

a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x\ne\pm1\end{matrix}\right.\)

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

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

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

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

b, Thay x = 0,5 vào A ta được : A = -3/2

Vậy ...

Answer 2

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

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

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

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

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

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

\(A=\left(\dfrac{1}{4}-1\right):\dfrac{1}{2}=\dfrac{-3}{4}\cdot2=-\dfrac{3}{2}\)

Vậy: Khi \(x=\dfrac{1}{2}\) thì \(A=-\dfrac{3}{2}\)