Question

Biến đổi mỗi biểu thức thành phân thức đại số :

a/ \(1-\dfrac{1}{1-\dfrac{1}{x}}\)

b/ \(\dfrac{\dfrac{x}{x+1}-\dfrac{x-1}{x}}{\dfrac{x}{x-1}-\dfrac{x-1}{x}}\)

Answer 1

a/ \(1-\dfrac{1}{1-\dfrac{1}{x}}=1-\dfrac{1}{\dfrac{x-1}{x}}=1-\dfrac{x}{x-1}=\dfrac{x-1-x}{x-1}=\dfrac{-1}{x-1}\)

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

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

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

Answer 2

a) \(1-\dfrac{1}{1-\dfrac{1}{x}}=1-\dfrac{1}{\dfrac{x-1}{x}}=\dfrac{x-1-x}{x-1}\)

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

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

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

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