Cho hàm số [f(x) = frac{{{x^2} + x - 1}}{{x - 1}} ]. Xét hai câu sau: [(I):f'(x) = 1 - frac{1}{{{{(x - 1)}^2}}}, ] [ forall x ne 1. ] [(II):f'(x) = frac{{{x^2} - 2x}}{{{{(x - 1)}^2}}}, ] [ forall x n...

Hướng dẫn giải: Chọn D Áp dụng công thức [{ left( { frac{u}{v}} right)^ prime } = frac{{u'.v - v'.u}}{{{v^2}}} ] ta có: [ forall x ne 1 ], ta có: [f(x) = frac{{{x^2} + x - 1}}{{x - 1}} ] [ Rightarrow ] [f'(x) = frac{{({x^2} + x - 1)'.(x - 1) - (x - 1)'.({x^2} + x - 1)}}{{{{(x - 1)}^2}}} ] [ Rightarrow ] [f'(x) ] [ = ] [ frac{{(2x + 1).(x - 1) - 1.({x^2} + x - 1)}}{{{{(x - 1)}^2}}} ] [ = ] [ frac{{2{x^2} - 2x + x - 1 - {x^2} - x + 1}}{{{{(x - 1)}^2}}} ] [ = ] [ frac{{{x^2} - 2x}}{{{{(x - 1)}^2}}} ] [ Rightarrow ] [(II) ]đúng. Mặt khác: [f'(x) ] [ = ] [ frac{{{x^2} - 2x}}{{{{(x - 1)}^2}}} = frac{{{x^2} - 2x + 1 - 1}}{{{{(x - 1)}^2}}} = frac{{{{(x - 1)}^2} - 1}}{{{{(x - 1)}^2}}} = 1 - frac{1}{{{{(x - 1)}^2}}} ] [ Rightarrow ] [(I) ]đúng.