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Câu hỏi: Cho hàm số $y=\dfrac{{{x}^{2}}-2x}{x+1}.$ Viết phương trình tiếp tuyến đồ thị hàm số tại điểm $A\left( 1;\dfrac{1}{2} \right).$
A. $y=\dfrac{1}{2}\left( x+1 \right)-\dfrac{1}{2}.$
B. $y=\dfrac{1}{4}\left( x+1 \right)+\dfrac{1}{2}.$
C. $y=\dfrac{1}{4}\left( x-1 \right)-\dfrac{1}{2}.$
D. $y=\dfrac{1}{2}\left( x-1 \right)+\dfrac{1}{2}.$
A. $y=\dfrac{1}{2}\left( x+1 \right)-\dfrac{1}{2}.$
B. $y=\dfrac{1}{4}\left( x+1 \right)+\dfrac{1}{2}.$
C. $y=\dfrac{1}{4}\left( x-1 \right)-\dfrac{1}{2}.$
D. $y=\dfrac{1}{2}\left( x-1 \right)+\dfrac{1}{2}.$
Ta có ${y}'=\dfrac{{{x}^{2}}+2x-2}{{{\left( x+1 \right)}^{2}}}\Rightarrow {y}'\left( 1 \right)=\dfrac{1}{4}\Rightarrow $ tiếp tuyến $y=\dfrac{1}{4}\left( x-1 \right)-\dfrac{1}{2}.$
Đáp án C.