\(\Delta=4-4\left(m-1\right)=8-4m\)
Để pt có 2 nghiệm phân biệt thì \(\Delta>0\Leftrightarrow m< 2\)
Theo hệ thức Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=2\\x_1x_2=m-1\end{matrix}\right.\)
Mặt khác: \(x_1^2-2x_2+x_1x_2=4\)
\(\Leftrightarrow x_1^2-\left(x_1+x_2\right)x_2+x_1x_2=4\) (vì \(x_1+x_2=2\))
\(\Leftrightarrow x_1^2-x_2^2=4\)
\(\Leftrightarrow\left(x_1-x_2\right)\left(x_1+x_2\right)=4\)
\(\Leftrightarrow x_1-x_2=2\) (vì \(x_1+x_2=2\))
Do đó: \(\left\{{}\begin{matrix}x_1+x_2=2\\x_1-x_2=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=2\\x_2=0\end{matrix}\right.\)
Mà \(x_1x_2=m-1\) nên \(m-1=0\Leftrightarrow m=1\) (tmđk)
Vậy m = 1 là giá trị cần tìm.
\(\text{#}Toru\)
\(\Delta=\left(-2\right)^2-4\left(m-1\right)=4-4m+4=-4m+8\)
Để phương trình có hai nghiệm phân biệt thì -4m+8>0
=>-4m>-8
=>m<2
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\\x_1x_2=\dfrac{c}{a}=m-1\end{matrix}\right.\)
\(x_1^2-2x_2+x_1x_2=4\)
=>\(x_1^2-x_2\left(x_1+x_2\right)+x_1x_2=4\)
=>\(x_1^2-x_2^2=4\)
=>\(\left(x_1+x_2\right)\left(x_1-x_2\right)=4\)
=>\(x_1-x_2=2\)
=>\(\left(x_1-x_2\right)^2=4\)
=>\(\left(x_1+x_2\right)^2-4x_1x_2=4\)
=>\(2^2-4\left(m-1\right)=4\)
=>4(m-1)=0
=>m-1=0
=>m=1(nhận)