a: Thay m=1 vào phương trình, ta được:
\(x^2-\left(2\cdot1-1\right)x+2\cdot1-4=0\)
=>\(x^2-x-2=0\)
=>(x-2)(x+1)=0
=>\(\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
b: \(\text{Δ}=\left[-\left(2m-1\right)\right]^2-4\cdot1\cdot\left(2m-4\right)\)
\(=\left(2m-1\right)^2-4\left(2m-4\right)\)
\(=4m^2-4m+1-8m+16\)
\(=4m^2-12m+17=4m^2-12m+9+8\)
\(=\left(2m-3\right)^2+8>=8>0\forall m\)
=>Phương trình luôn có hai nghiệm phân biệt
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{-\left[-\left(2m-1\right)\right]}{1}=2m-1\\x_1\cdot x_2=\dfrac{c}{a}=\dfrac{2m-4}{1}=2m-4\end{matrix}\right.\)
Theo đề, ta có hệ phương trình:
\(\left\{{}\begin{matrix}x_1+x_2=2m-1\\x_1-2x_2=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x_2=2m-1-3=2m-4\\x_1+x_2=2m-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x_2=\dfrac{2}{3}m-\dfrac{4}{3}\\x_1=2m-1-\dfrac{2}{3}m+\dfrac{4}{3}=\dfrac{4}{3}m+\dfrac{1}{3}\end{matrix}\right.\)
\(x_1\cdot x_2=2m-4\)
=>\(\left(\dfrac{2}{3}m-\dfrac{4}{3}\right)\left(\dfrac{4}{3}m+\dfrac{1}{3}\right)=2m-4\)
=>\(\dfrac{1}{9}\left(2m-4\right)\left(4m+1\right)=2m-4\)
=>\(\left(2m-4\right)\left(4m+1\right)=18m-36\)
=>\(\left(m-2\right)\left(8m+2\right)-18\left(m-2\right)=0\)
=>\(\left(m-2\right)\left(8m+2-18\right)=0\)
=>\(\left(m-2\right)\left(8m-16\right)=0\)
=>\(8\left(m-2\right)^2=0\)
=>\(\left(m-2\right)^2=0\)
=>m-2=0
=>m=2(nhận)
c:
\(x_1^2\cdot x_2+x_1\cdot x_2^2+3\left(x_1+x_2\right)=0\)
=>\(x_1x_2\left(x_1+x_2\right)+3\left(x_1+x_2\right)=0\)
=>\(\left(x_1+x_2\right)\left(x_1x_2+3\right)=0\)
=>\(\left(2m-1\right)\left(2m-4+3\right)=0\)
=>\(\left(2m-1\right)^2=0\)
=>2m-1=0
=>2m=1
=>\(m=\dfrac{1}{2}\)
d: \(A=x_1^2+x_2^2\)
\(=\left(x_1+x_2\right)^2-2x_1x_2\)
\(=\left(2m-1\right)^2-2\left(2m-4\right)\)
\(=4m^2-4m+1-4m+8\)
\(=4m^2-8m+9\)
\(=4m^2-8m+4+5=\left(2m-2\right)^2+5>=5\forall m\)
Dấu '=' xảy ra khi 2m-2=0
=>2m=2
=>m=1
e: \(\left\{{}\begin{matrix}x_1+x_2=2m-1\\x_1\cdot x_2=2m-4\end{matrix}\right.\)
=>\(x_1+x_2-x_1x_2=2m-1-\left(2m-4\right)=2m-1-2m+4=3\)
f: \(\dfrac{1}{x_1}+\dfrac{1}{x_2}>=1\)
=>\(\dfrac{x_1+x_2}{x_1x_2}>=1\)
=>\(\dfrac{2m-1}{2m-4}-1>=0\)
=>\(\dfrac{2m-1-2m+4}{2m-4}>=0\)
=>\(\dfrac{3}{2m-4}>=0\)
=>2m-4>0
=>2m>4
=>m>2