a: Thay m=2 vào (1), ta được:
\(x^2-2\left(2+1\right)x+4\cdot2=0\)
=>x^2-6x+8=0
=>(x-2)(x-4)=0
=>\(\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
b: \(\Delta=\left[-2\left(m+1\right)\right]^2-4\cdot1\cdot4m\)
\(=4m^2+8m+4-16m=4m^2-8m+4=\left(2m-2\right)^2\)
Để phương trình (1) có hai nghiệm phân biệt thì \(\Delta>0\)
=>\(\left(2m-2\right)^2>0\)
=>\(2m-2\ne0\)
=>m<>1
Theo vi-et, ta có;
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\left(m+1\right)\\x_1x_2=\dfrac{c}{a}=4m\end{matrix}\right.\)
\(\left|x_1\right|-\left|x_2\right|=-4\)
=>\(\left(\left|x_1\right|-\left|x_2\right|\right)^2=16\)
=>\(x_1^2+x_2^2-2\left|x_1x_2\right|=16\)
=>\(\left(x_1+x_2\right)^2-2x_1x_2-2\left|x_1x_2\right|=16\)
=>\(\left(2m+2\right)^2-2\cdot4m+2\left|4m\right|=16\)
=>\(4m^2+8m+4-8m+8\left|m\right|-16=0\)
=>\(4m^2-12+8\left|m\right|=0\)
=>\(m^2+2\left|m\right|-3=0\)
=>\(\left(\left|m\right|+3\right)\left(\left|m\right|-1\right)=0\)
=>|m|-1=0
=>m=1(loại) hoặc m=-1(nhận)
