Question

\(\sqrt[3]{x^4+x-1}+x^2-1\ge0\)

Answer 1

\(\sqrt{x^4+x^2-1}+x^2-1\ge0\)

=>\(\sqrt{x^4+x^2-1}\ge-x^2+1\) (1)

TH1; \(-x^2+1<0\)

=>\(-x^2<-1\)

=>\(x^2>1\)

=>x>1 hoặc x<-1

\(\sqrt{x^4+x^2-1}\ge-x^2+1\)

mà \(\sqrt{x^4+x^2-1}\ge0;-x^2+1<0\)

nên x>1 hoặc x<-1 thỏa mãn

TH2: \(-x^2+1\ge0\)

=>\(-x^2\ge-1\)

=>\(x^2\le1\)

=>-1<=x<=1

\(\sqrt{x^4+x^2-1}\ge-x^2+1\)

=>\(x^4+x^2-1\ge\left(-x^2+1\right)^2=x^4-2x^2+1\)

=>\(3x^2\ge2\)

=>\(x^2\ge\frac23\)

=>\(\left[\begin{array}{l}x\ge\frac{\sqrt6}{3}\\ x\le-\frac{\sqrt6}{3}\end{array}\right.\)

mà -1<=x<=1

nên \(-1\le x\le-\frac{\sqrt6}{3}\) hoặc \(\frac{\sqrt6}{3}\) <=x<=1