\(x^2-\left(m+2\right)x+m=0\) (1)
a) Ta có: \(\Delta=\left[-\left(m+2\right)\right]^2-4\cdot1\cdot m=m^2+4m+4-4m=m^2+4>0\forall m\in R\)
\(\Rightarrow\) PT(1) luôn có 2 nghiệm pb x1, x2 với mọi m
b) Theo ĐL Vi-ét ta có: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=m+2\\x_1x_2=\dfrac{c}{a}=m\end{matrix}\right.\)
Ta có: \(\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{1}{x_1+x_2-2}\)
\(\Leftrightarrow\dfrac{x_1+x_2}{x_1x_2}=\dfrac{1}{x_1+x_2-2}\)
\(\Leftrightarrow\dfrac{m+2}{m}=\dfrac{1}{m+2-2}\)
\(\Leftrightarrow\dfrac{m+2}{m}=\dfrac{1}{m}\) (ĐK: \(m\ne0\) )
\(\Leftrightarrow m\left(m+2\right)=m\)
\(\Leftrightarrow m^2+2m=m\)
\(\Leftrightarrow m^2+m=0\)
\(\Leftrightarrow m\left(m+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}m=0\left(L\right)\\m=-1\end{matrix}\right.\)
Vậy m = -1 là GT cần tìm
