Nếu $f\left(x\right)=\dfrac{x^2-2x+5}{x-1}$ thì $f'\left(2\right)$ bằng
![]() | $-3$ |
![]() | $-5$ |
![]() | $0$ |
![]() | $1$ |
Chọn phương án A.
$\begin{aligned}
f'\left(x\right)&=\dfrac{\left(2x-2\right)\left(x-1\right)-\left(x^2-2x+5\right)}{\left(x-1\right)^2}\\
&=\dfrac{x^2-2x-3}{\left(x-1\right)^2}.
\end{aligned}$
Do đó $f'\left(2\right)=\dfrac{2^2-2.2-3}{\left(2-1\right)^2}=-3$.