Đáp án:
$0<x<1$ hoặc $x<-1$
Giải thích các bước giải:
Ta có :
$3(2x^2-x\sqrt{x^2+3})<2(1-x^4)$
$\rightarrow 3\dfrac{(2x^2-x\sqrt{x^2+3})(2x^2-x\sqrt{x^2+3}))}{2x^2+x\sqrt{x^2+3})}<2(1-x^4)$
$\rightarrow 3\dfrac{4x^4-x^2(x^2+3)}{2x^2+x\sqrt{x^2+3})}<2(1-x^4)$
$\rightarrow 0<2(1-x^4)+ 3\dfrac{3(1-x^4)}{2x^2+x\sqrt{x^2+3}}$
$\rightarrow 0<(1-x^4)(2+ 3\dfrac{3}{2x^2+x\sqrt{x^2+3}})$
$\rightarrow 0<(1-x^4)(\dfrac{4x^2+2x\sqrt{x^2+3}+9}{2x^2+x\sqrt{x^2+3}})$
$\rightarrow 0<(1-x^4)(\dfrac{x^2+2x\sqrt{x^2+3}+3(x^2+3)}{2x^2+x\sqrt{x^2+3}})$
$\rightarrow 0<(1-x^4).\dfrac{1}{x(2x+\sqrt{x^2+3})}$
$+)1-x^4>0\rightarrow 1>x^4\rightarrow -1<x<1$
$\rightarrow x(2x+\sqrt{x^2+3})>0\rightarrow x>0$
Hoặc $x<0, 2x+\sqrt{x^2+3}<0$
$\rightarrow \sqrt{x^2+3}<-2x\rightarrow x^2+3<4x^2\rightarrow x^2>1\rightarrow x<-1$ do $x<0$
$\rightarrow $ Loại do $-1<x<1$
$\Rightarrow 0<x<1$
$+)1-x^4<0\rightarrow x>1$ hoặc $x<-1$
Nếu $x>1\rightarrow x(2x+\sqrt{x^2+3})>0\rightarrow (1-x^4).x(2x+\sqrt{x^2+3})<0\rightarrow$ Loại
Nếu $x<-1\rightarrow x(2x+\sqrt{x^2+3})>0$
$\rightarrow 2x+\sqrt{x^2+3}<0$
$\rightarrow \sqrt{x^2+3}<-2x$
$\rightarrow x^2+3<4x^2$
$\rightarrow x^2>1\rightarrow x<-1$
$\Rightarrow x<-1$
