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Câu hỏi: Viết phương trình tiếp tuyến của đồ thị hàm số $y=f\left( x \right)$ tại điểm có hoành độ $x=1$ và thỏa mãn $f\left( 1+x \right)=x-{{f}^{2}}\left( 1-x \right).$
A. $y=\dfrac{1}{3}x-\dfrac{4}{3}$ hoặc $y=x-1.$
B. $y=\dfrac{1}{3}x-\dfrac{4}{3}$ hoặc $y=x+1.$
C. $y=\dfrac{1}{3}x+\dfrac{4}{3}$ hoặc $y=x-1.$
D. $y=\dfrac{1}{3}x+\dfrac{4}{3}$ hoặc $y=x+1.$
A. $y=\dfrac{1}{3}x-\dfrac{4}{3}$ hoặc $y=x-1.$
B. $y=\dfrac{1}{3}x-\dfrac{4}{3}$ hoặc $y=x+1.$
C. $y=\dfrac{1}{3}x+\dfrac{4}{3}$ hoặc $y=x-1.$
D. $y=\dfrac{1}{3}x+\dfrac{4}{3}$ hoặc $y=x+1.$
Từ $f\left( 1+x \right)=x-{{f}^{2}}\left( 1-x \right)\Rightarrow f\left( 1 \right)=0-{{f}^{2}}\left( 1 \right)\Rightarrow \left[ \begin{array}{*{35}{l}}
f\left( 1 \right)=0 \\
f\left( 1 \right)=-1 \\
\end{array} \right.$
Từ $f\left( 1+x \right)=x-{{f}^{2}}\left( 1-x \right)\Rightarrow {f}'\left( 1+x \right)=1+2f\left( 1-x \right).{f}'\left( 1-x \right)\Rightarrow {f}'\left( 1 \right)=1+2f\left( 1 \right).{f}'\left( 1 \right).$
$f\left( 1 \right)=0\Rightarrow {f}'\left( 1 \right)=1\Rightarrow d:y={f}'\left( 1 \right).\left( x-1 \right)+f\left( 1 \right)\Rightarrow d:y=x-1.$
$f\left( 1 \right)=-1\Rightarrow {f}'\left( 1 \right)=1-2{f}'\left( 1 \right)\Rightarrow {f}'\left( 1 \right)=\dfrac{1}{3}$
$\Rightarrow d:y={f}'\left( 1 \right).\left( x-1 \right)+f\left( 1 \right)\Rightarrow d:y=\dfrac{1}{3}\left( x-1 \right)-1\Leftrightarrow y=\dfrac{1}{3}x-\dfrac{4}{3}.$
f\left( 1 \right)=0 \\
f\left( 1 \right)=-1 \\
\end{array} \right.$
Từ $f\left( 1+x \right)=x-{{f}^{2}}\left( 1-x \right)\Rightarrow {f}'\left( 1+x \right)=1+2f\left( 1-x \right).{f}'\left( 1-x \right)\Rightarrow {f}'\left( 1 \right)=1+2f\left( 1 \right).{f}'\left( 1 \right).$
$f\left( 1 \right)=0\Rightarrow {f}'\left( 1 \right)=1\Rightarrow d:y={f}'\left( 1 \right).\left( x-1 \right)+f\left( 1 \right)\Rightarrow d:y=x-1.$
$f\left( 1 \right)=-1\Rightarrow {f}'\left( 1 \right)=1-2{f}'\left( 1 \right)\Rightarrow {f}'\left( 1 \right)=\dfrac{1}{3}$
$\Rightarrow d:y={f}'\left( 1 \right).\left( x-1 \right)+f\left( 1 \right)\Rightarrow d:y=\dfrac{1}{3}\left( x-1 \right)-1\Leftrightarrow y=\dfrac{1}{3}x-\dfrac{4}{3}.$
Đáp án A.