\(y'=1-\dfrac{1}{\left(x-1\right)^2}\)
Tiếp tuyến d tại \(A\left(a;a+1+\dfrac{1}{a-1}\right)\) có dạng:
\(y=\left(1-\dfrac{1}{\left(a-1\right)^2}\right)\left(x-a\right)+a+1+\dfrac{1}{a-1}\)
\(\Leftrightarrow y=\dfrac{a^2-2a}{\left(a-1\right)^2}x+\dfrac{a^2}{\left(a-1\right)^2}\)
\(\Rightarrow M\left(\dfrac{a^2}{2a-a^2};0\right)\) ; \(N\left(0;\dfrac{a^2}{\left(a-1\right)^2}\right)\Rightarrow\left\{{}\begin{matrix}OM=\dfrac{a^2}{\left|2a-a^2\right|}\\ON=\dfrac{a^2}{\left(a-1\right)^2}\end{matrix}\right.\)
\(\dfrac{a^2}{\left(a-1\right)^2}=\dfrac{2a^2}{\left|2a-a^2\right|}\Leftrightarrow\left[{}\begin{matrix}a=0\left(loại\right)\\\left|a^2-2a\right|=2\left(a^2-2a+1\right)\end{matrix}\right.\)
Đặt \(a^2-2a=t\Rightarrow\left|t\right|=2\left(t+1\right)\) (với \(t\ge-1\))
\(\Leftrightarrow\left[{}\begin{matrix}2t+2=t\\2t+2=-t\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}t=-2\left(loại\right)\\t=-\dfrac{2}{3}\end{matrix}\right.\)
\(\Rightarrow a^2-2a=-\dfrac{2}{3}\Leftrightarrow a^2-2a+\dfrac{2}{3}=0\)
Người ra đề đam mê với nghiệm xấu thì phải
