Question

Câu 2: Cho phương trình -x2 +2(2+m)x-m2=0 Co tham so m) |x1+x2-4|=2 x1 x2 Giup mk nha cam on

Answer 1

\(-x^2+2\left(2+m\right)x-m^2=0\)

Theo Vi - ét, ta có :

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\left(2+m\right)\\x_1x_2=\dfrac{c}{a}=m^2\end{matrix}\right.\)

Ta có :

\(\left|x_1+x_2-4\right|=2x_1x_2\)

\(\Leftrightarrow\left|4+2m-4\right|=2m^2\)

\(\Leftrightarrow\left|2m\right|=2m^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2m=2m^2\\2m=-2m^2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2m-2m^2=0\\2m+2m^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2m\left(1-m\right)=0\\2m\left(1+m\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}m=0\\m=1\\m=-1\end{matrix}\right.\)