pt có 2 no
\(\Delta=\left(-2\right)^2-4\left(2-m\right)=4-8+4m\\ =4m-4\)
\(\Delta\ge0\\ 4m-4\ge0\\ m\ge1\)
ÁP dụng Vi ét
\(\left\{{}\begin{matrix}x_1+x_2=2\\x_1x_2=2-m\end{matrix}\right.\\ \left(\left|x_1-x_2\right|\right)^2=4\\ \left(x_1-x_2\right)^2=4\\ x^2_1-2x_1x_2+x_2^2=4\\ \left(x_1+x_2\right)^2-4x_1x_2=4\\ 2^2-4\left(2-m\right)=4\\ 4-4\left(2-m\right)=4\\ 4\left(2-m\right)=0\\ 2-m=0\\ m=2\left(thoamanđk\right)\)
\(\text{Δ}=2^2-4\cdot\left(2-m\right)=4-8+4m=4m-4\)
Đểpt có hai nghiệm thì 4m-4>=0
hay m>=1
Theo đề, ta có:
\(\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}=2\)
\(\Leftrightarrow\sqrt{2^2-4\left(2-m\right)}=2\)
=>4-8+4m=4
=>4m-4=4
hay m=2
