\(\left(x^2-2x-3\right)^2\ge\left(x^2+3x+3\right)^2\)
=>\(\left(x^2-2x-3\right)^2-\left(x^2+3x+3\right)^2\ge0\)
=>\(\left(x^2-2x-3-x^2-3x-3\right)\left(x^2-2x-3+x^2+3x+3\right)\) >=0
=>(-5x-6)(2x^2+x)>=0
=>x(2x+1)(5x+6)<=0
Đặt f(x)=x(2x+1)(5x+6)
Đặt x=0
=>x=0
Đặt 2x+1=0
=>x=-1/2
Đặt 5x+6=0
=>x=-6/5
Bảng xét dấu:
f(x)<=0
=>x<=-6/5; -1/2<=x<=0