a: \(x^2-2\left|x\right|-3=0\)
=>\(\left(\left|x\right|\right)^2-3\cdot\left|x\right|+\left|x\right|-3=0\)
=>\(\left(\left|x\right|-3\right)\left(\left|x\right|+1\right)=0\)
=>|x|-3=0
=>|x|=3
=>x=3 hoặc x=-3
b: Ta có: \(x\cdot\left|x+3\right|-\left|x^2+x+1\right|=1\)
=>\(x\cdot\left|x+3\right|-x^2-x-1=1\)
=>\(x\cdot\left|x+3\right|-x^2-x-2=0\) (1)
TH1: x>=-3
(1) sẽ trở thành:
\(x\left(x+3\right)-x^2-x-2=0\)
=>\(x^2+3x-x^2-x-2=0\)
=>2x-2=0
=>2x=2
=>x=1(nhận)
TH2: x<-3
(1) sẽ trở thành: \(x\left(-x-3\right)-x^2-x-2=0\)
=>\(-x^2-3x-x^2-x-2=0\)
=>\(2x^2+4x+2=0\)
=>\(x^2+2x+1=0\)
=>\(\left(x+1\right)^2=0\)
=>x+1=0
=>x=-1(loại)
