a: Thay x=3 vào phương trình, ta được:
\(3^2-2\left(m-1\right)\cdot3+2m-5=0\)
=>\(9-6\left(m-1\right)+2m-5=0\)
=>2m+4-6m+6=0
=>-4m=-10
=>\(m=\dfrac{5}{2}\)
Theo Vi-et, ta có:
\(x_1+x_2=\dfrac{-b}{a}=2\left(m-1\right)=2\left(\dfrac{5}{2}-1\right)=3\)
=>\(x_2=3-x_1=3-3=0\)
b: \(x^2-2\left(m-1\right)x+2m-5=0\)
\(\text{Δ}=\left[-2\left(m-1\right)\right]^2-4\cdot1\cdot\left(2m-5\right)\)
\(=4\left(m^2-2m+1\right)-4\left(2m-5\right)\)
\(=4\left(m^2-4m+6\right)\)
\(=4\left(m^2-4m+4+2\right)\)
\(=4\left(m-2\right)^2+8>0\forall m\)
=>Phương trình luôn có hai nghiệm phân biệt
Theo vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\left(m-1\right)=2m-2\\x_1x_2=\dfrac{c}{a}=2m-5\end{matrix}\right.\)
\(x_1^2-2m\cdot x_1+2m-1\)
\(=x_1^2+x_1\left(-2m+2\right)-2x_1+2m-5+4=0-2x_1+4=-2x_1+4\)
\(\left(x_1^2-2mx_1+2m-1\right)\left(2-x_2\right)=2\)
=>\(\left(-2x_1+4\right)\left(2-x_2\right)=2\)
=>\(-2\left(x_1-2\right)\cdot\left(-1\right)\cdot\left(x_2-2\right)=2\)
=>\(\left(x_1-2\right)\left(x_2-2\right)=1\)
=>\(x_1x_2-2\left(x_1+x_2\right)+4=1\)
=>2m-5-2(2m-2)+4=1
=>2m-5-4m+4+4=1
=>-2m+3=1
=>-2m=-2
=>m=1
\(\left(x_1^2-2mx_1+2m-1\right)\left(2-x_2\right)=2\)
=>
