a: Thay x=2 vào phương trình, ta được:
\(2^2-2\left(m+1\right)\cdot2+2m-3=0\)
=>\(4-4\left(m+1\right)+2m-3=0\)
=>4-4m-4+2m-3=0
=>-2m-3=0
=>\(m=-\dfrac{3}{2}\)
\(x_1+x_2=-\dfrac{b}{a}=2\left(m+1\right)=2\left(-\dfrac{3}{2}+1\right)=-1\)
=>\(x_2+2=-1\)
=>\(x_2=-3\)
b:
\(\Delta=\left[-2\left(m+1\right)\right]^2-4\left(2m-3\right)\)
\(=4m^2+8m+4-8m+12\)
\(=4m^2+16>0\forall m\)
=>Phương trình luôn có hai nghiệm phân biệt
x1,x2 là các nghiệm của phương trình
=>\(\left\{{}\begin{matrix}x_1^2-2\left(m+1\right)x_1+2m-3=0\\x_2^2-2\left(m+1\right)x_2+2m-3=0\end{matrix}\right.\)
\(x_1^2-\left(2m+1\right)x_1+2m-1\)
\(=x_1^2-\left(2m+2\right)x_1+x_1+2m-3+2\)
\(=0+2+x_1=x_1+2\)
\(x_2^2-\left(2m+1\right)x_2+2m-1\)
\(=x_2^2-\left(2m+2\right)x_2+2m-3+x_2+2\)
\(=x_2+2\)
Theo vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\left(m+1\right)=2m+2\\x_1x_2=\dfrac{c}{a}=2m-3\end{matrix}\right.\)
\(\left[x_1^2-\left(2m+1\right)x_1+2m-1\right]\left[x_2^2-\left(2m+1\right)x_2+2m-1\right]=5\)
=>\(\left(x_1+2\right)\left(x_2+2\right)=5\)
=>\(x_1x_2+2\left(x_1+x_2\right)+4=5\)
=>\(2m-3+2\left(2m+2\right)+4=5\)
=>2m+1+4m+4=5
=>6m=5-3=2
=>\(m=\dfrac{1}{3}\left(nhận\right)\)
