a) \(x^2-\left(m+2\right)x+2m=0\) (1)
Thay m = -1 vào (1), ta được:
\(x^2-\left(-1+2\right)x+2\cdot\left(-1\right)=0\)
\(\Leftrightarrow x^2-x-2=0\)
\(\Leftrightarrow x^2+x-2x-2=0\)
\(\Leftrightarrow x\left(x+1\right)-2\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
b) \(\Delta=\left(m+2\right)^2-4\cdot2m=\left(m-2\right)^2\ge0;\forall m\)
\(\Rightarrow\) Phương trình (1) có hai nghiệm
Theo hệ thức Vi-ét: \(\left\{{}\begin{matrix}x_1+x_2=m+2\\x_1x_2=2m\end{matrix}\right.\)
Lại có: \(\dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}=2\)
\(\Leftrightarrow\dfrac{x_1^2+x_2^2}{x_1x_2}=2\)
\(\Rightarrow\left(x_1+x_2\right)^2-2x_1x_2=2x_1x_2\)
\(\Leftrightarrow\left(m+2\right)^2-2\cdot2m=2\cdot2m\)
\(\Leftrightarrow m^2+4m+4-8m=0\)
\(\Leftrightarrow m^2-4m+4=0\)
\(\Leftrightarrow\left(m-2\right)^2=0\)
\(\Leftrightarrow m-2=0\)
\(\Leftrightarrow m=2\)
$\text{#}Toru$
