a/ Tạo với 2 trục tọa độ một tam giác vuông cân, tức là hệ số góc của tiếp tuyến bằng \(\pm1\). Hay \(f'\left(x\right)=\pm1\)
\(f'\left(x\right)=\dfrac{x-1-x}{\left(x-1\right)^2}=-\dfrac{1}{\left(x-1\right)^2}\)
\(\left(x-1\right)^2>0\forall x\ne1\Rightarrow f'\left(x\right)=-1\)
\(\Leftrightarrow x-1=\pm1\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=2\\y=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}y=-1\left(x-2\right)+2=4-x\\y=-1\left(x-0\right)+0=-x\end{matrix}\right.\)
b/ \(y=k\left(x-1\right)+3\)
\(\left\{{}\begin{matrix}k\left(x-1\right)+3=\dfrac{x}{x-1}\left(1\right)\\k=-\dfrac{1}{\left(x-1\right)^2}\left(2\right)\end{matrix}\right.\)
The (2) vo (1) \(\Rightarrow-\dfrac{x-1}{\left(x-1\right)^2}+3=\dfrac{x}{x-1}\Leftrightarrow\dfrac{-1}{x-1}+3=\dfrac{x}{x-1}\)
\(\Leftrightarrow\dfrac{x+1}{x-1}=3\Leftrightarrow x+1=3x-3\Leftrightarrow x=2\)
\(\Rightarrow k=-\dfrac{1}{\left(2-1\right)^2}=-1;y=2\)
\(\Rightarrow y=-1\left(x-2\right)+2=4-x\)
P/s: Check lại dùm toi nha
