\(y=\dfrac{2x+2}{x-1}\Rightarrow y'=\dfrac{2\left[x-1-\left(x+1\right)\right]}{\left(x-1\right)^2}=\dfrac{-4}{\left(x-1\right)^2}\)
Giả sử d là tiếp tuyến của (C) tại \(M\left(x_o;y_o\right)\)
Khi đó : PTTT d : \(y=\dfrac{-4}{\left(x_o-1\right)^2}\left(x-x_o\right)+\dfrac{2x_o+2}{x_o-1}\)
\(\Rightarrow y+\dfrac{4}{\left(x_o-1\right)^2}x-\dfrac{4x_o+2\left(x_o^2-1\right)}{\left(x_o-1\right)^2}=0\)
Ta có : d(I;d) = \(\left|\dfrac{\dfrac{4}{\left(x_o-1\right)^2}.1+1.2-\dfrac{4x_o+2x_o^2-2}{\left(x_o-1\right)^2}}{\sqrt{\dfrac{16}{\left(x_o-1\right)^4}+1}}\right|\)
\(=\left|\dfrac{4+2\left(x_o-1\right)^2-4x_o-2x_o^2+2}{\sqrt{16+\left(x_o-1\right)^4}}\right|\)
\(=\left|\dfrac{8\left(1-x_o\right)}{\sqrt{16+\left(1-x_o\right)^4}}\right|\le8\left|\dfrac{\left(1-x_o\right)}{\sqrt{8\left(1-x_o\right)^2}}\right|=\sqrt{8}\)
" = " \(\Leftrightarrow\left[{}\begin{matrix}1-x_o=2\\1-x_o=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x_o=-1\\x_o=3\end{matrix}\right.\)
Với xo = -1 . Suy ra : \(y=-\left(x+1\right)=-x-1\)
Với xo = 3 . Suy ra : \(y=-\left(x-3\right)+4=-x+7\)
