HS có 1 \(TCN:y=0\left(\lim\limits_{x\rightarrow\infty}y=0\right)\)
Để HS có 2 tiệm cận khi \(g\left(x\right)=x^2+2\left(m-2\right)x+m^2+1=0\) có nghiệm kép hoặc có 1 nghiệm \(x=2\)
\(TH1:\Delta'=\left(m-2\right)^2-m^2-1=0\)
\(\Leftrightarrow-4m=-3\)
\(\Leftrightarrow m=\dfrac{3}{4}\left(1\right)\)
\(TH2:\left\{{}\begin{matrix}\Delta'=-4m+3>0\\g\left(2\right)=4+4\left(m-2\right)+m^2+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m< \dfrac{3}{4}\\m^2+4m-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m< \dfrac{3}{4}\\m=-2\pm\sqrt{7}\left(tm\right)\end{matrix}\right.\) \(\Leftrightarrow m=-2\pm\sqrt{7}\left(2\right)\)
\(\left(1\right);\left(2\right)\Rightarrow\)Tích các giá trị của \(m\) là :\(\dfrac{3}{4}.\left(-2-\sqrt{7}\right)\left(-2+\sqrt{7}\right)=\dfrac{3}{4}.\left(-3\right)=\dfrac{-9}{4}=\dfrac{a}{b}\Rightarrow\left\{{}\begin{matrix}a=-9\\b=4\end{matrix}\right.\)
\(P=a^2+b^2=81+16=97\)
