1: Thay m=0 vào (1), ta được:
\(x^2+2\left(0+1\right)x-0^2+2\cdot0-3=0\)
=>\(x^2+2x-3=0\)
=>(x+3)(x-1)=0
=>\(\left[\begin{array}{l}x+3=0\\ x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-3\\ x=1\end{array}\right.\)
2: \(\Delta=\left(2m+2\right)^2-4\cdot1\cdot\left(-m^2+2m-3\right)\)
\(=4m^2+8m+4+4m^2-8m+12=8m^2+16\) >0∀m
=>Phương trình luôn có hai nghiệm phân biệt
Theo Vi-et, ta có: \(\begin{cases}x_1+x_2=-\frac{b}{a}=-2\left(m+1\right)\\ x_1x_2=\frac{c}{a}=-m^2+2m-3\end{cases}\)
\(\left(x_1+x_2+x_1x_2\right)^2\)
\(=\left(-2m-2-m^2+2m-3\right)^2=\left(-m^2-5\right)^2=\left(m^2+5\right)^2\)
x1 là nghiệm của phương trình nên ta có:
\(x_1^2+2\left(m+1\right)x_1-m^2+2m-3=0\)
=>\(x_1^2+2\left(m+1\right)x_1+\left(m+1\right)^2-\left(m+1\right)^2-m^2+2m-3=0\)
=>\(\left(x_1+m+1\right)^2-m^2-2m-1-m^2+2m-3=0\)
=>\(\left(x_1+m+1\right)^2-2m^2-4=0\)
=>\(\left(x_1+m+1\right)^2=2m^2+4\)
=>\(\sqrt{\left(x_1+m+1\right)^2}=\sqrt{2m^2+4}\)
=>\(\left|x_1+m+1\right|=\sqrt{2\left(m^2+2\right)}\)
Ta có: \(7\cdot\left|x_1+m+1\right|\cdot\sqrt{2m^2+4}-6=\left(x_1+x_2+x_1x_2\right)^2\)
=>\(7\cdot\sqrt{2\left(m^2+2\right)}\cdot\sqrt{2\left(m^2+2\right)}-6=\left(m^2+5\right)^2\)
=>\(7\cdot2\cdot\left(m^2+2\right)-6=\left(m^2+5\right)^2\)
=>\(m^4+10m^2+25=14m^2+28-6=14m^2+22\)
=>\(m^4-4m^2+3=0\)
=>\(\left(m^2-1\right)\left(m^2-3\right)=0\)
=>\(\left[\begin{array}{l}m^2-1=0\\ m^2-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}m^2=1\\ m^2=3\end{array}\right.\Rightarrow\left[\begin{array}{l}m\in\left\lbrace1;-1\right\rbrace\\ m\in\left\lbrace\sqrt3;-\sqrt3\right\rbrace\end{array}\right.\)
