Phương trình hoành độ giao điểm là:
\(x^2=4x-2m+1\)
=>\(x^2-4x+2m-1=0\)
\(\Delta=\left(-4\right)^2-4\left(2m-1\right)=16-8m+4=-8m+20\)
Để phương trình có hai nghiệm phân biệt thì -8m+20>0
=>-8m>-20
=>m<2,5
Theo Vi-et, ta có: \(\begin{cases}x_1+x_2=-\frac{b}{a}=4\\ x_1x_2=\frac{c}{a}=2m-1\end{cases}\)
\(4x_1x_2=4\left(2m-1\right)\)
\(\left(\left|x_1\right|+\left|x_2\right|\right)^2=x_1^2+x_2^2+2\cdot\left|x_1x_2\right|\)
\(=\left(x_1+x_2\right)^2-2\cdot x_1\cdot x_2+2\cdot\left|x_1x_2\right|\)
\(=4^2-2\left(2m-1\right)+2\left|2m-1\right|=16-2\left(2m-1\right)+2\left|2m-1\right|\)
TH1: \(m\ge\frac12\)
=>\(\left(\left|x_1\right|+\left|x_2\right|\right)^2=16-2\left(2m-1\right)+2\left(2m-1\right)=16\)
=>\(\left|x_1\right|+\left|x_2\right|=4\)
Ta có: \(\left|x_1\right|+\left|x_2\right|+4x_1x_2\ge10\)
=>4+4(2m-1)>=10
=>4(2m-1)>=6
=>2m-1>=3/2
=>\(2m\ge\frac52\)
=>\(m\ge\frac54\) (nhận)
TH2: \(m<\frac12\)
\(\left(\left|x_1\right|+\left|x_2\right|\right)^2=16-2\left(2m-1\right)+2\left|2m-1\right|\)
\(=16-2\left(2m-1\right)-2\left(2m-1\right)=16-4\left(2m-1\right)\)
=>\(\left|x_1\right|+\left|x_2\right|=\sqrt{16-4\left(2m-1\right)}=\sqrt{4\cdot\left(4-2m+1\right)}=2\cdot\sqrt{5-2m}\)
Ta có: \(\left|x_1\right|+\left|x_2\right|+4x_1x_2\ge10\)
=>\(2\cdot\sqrt{5-2m}\) +4(2m-1)>=10
=>\(\sqrt{5-2m}+2\left(2m-1\right)\ge10\)
=>\(\sqrt{5-2m}+4m-2-10\ge0\)
=>\(\sqrt{5-2m}\ge-4m+12\)
TH1: -4m+12<=0
=>-4m<=-12
=>m>=3(loại)
TH2: -4m+12>=0
=>-4m>=-12
=>m<=3
=>\(\frac12\le m\le3\)
Ta có: \(\sqrt{5-2m}\ge-4m+12\)
=>\(5-2m\ge\left(-4m+12\right)^2\)
=>\(16m^2-96m+144+2m-5\le0\)
=>\(16m^2-94m+139\le0\) (1)
\(\Delta=\left(-94\right)^2-4\cdot16\cdot139=8836-8896=-60<0\)
Vì Δ<0 và a=16>0
nên \(16m^2-94m+139>0\forall m\)
=>(1) vô nghiệm
Vậy: \(m\ge\frac54\)
2 =3