Cho hàm số: 

A

- Hướng dẫn giải

Với ∆x là số gia của đối số tại x=1, ta có

$\begin{array}{l}\Delta y=\frac{{(1+\text{}\Delta x)}^2-2(\text{}1+\text{}\Delta x)\text{}}{1+\text{}\Delta x+\text{}1}-\frac{1-2}{1+\text{}1}\\ =\frac{1+\text{}2\Delta x+\text{}{\left(\Delta x\right)}^2-2-2\Delta x}{2+\text{}\Delta x}+\text{}\frac{1}{2}\\ =\frac{{\left(\Delta x\right)}^2-1}{2+\text{}\Delta x}\text{\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}\hspace{0.17em}}\frac{1}{2}=\frac{2{\left(\Delta x\right)}^2-2+\text{}2+\text{}\Delta x}{2(\text{}2+\text{}\Delta x)}\\ =\frac{2{\left(\Delta x\right)}^2+\text{}\Delta x}{2(2+\text{}\Delta x)}=\frac{(2\Delta x+1)\text{}.\text{}\Delta x}{2(2+\text{}\Delta x)}\\ \frac{\Delta y}{\Delta x}=\frac{2\Delta x+1}{2(2+\text{}\Delta x)}\end{array}$

Vậy y’(1)$= \underset{∆x\to 0}{lim}\frac{∆y}{∆x}= \frac{1}{4}$

 Đáp án A