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PT
$\Leftrightarrow x^{2} - 2x-1= \sqrt[3]{x^{4} - x^{2} } $
$\Leftrightarrow (x^{2} - 2x-1)^3=x^{4} - x^{2} $
$\Leftrightarrow x^6 -6x^5+10x^4+4x^3-10x^2-6x-1=0 $
$\Leftrightarrow (x^{2} - x-1)( x^4-5x^3+6x^2+5x+1)=0 $
Pt $x^{2} - x-1$ cho nghiệm $x=\dfrac{1\pm \sqrt 5}{2}$
PT $ x^4-5x^3+6x^2+5x+1=0$ vô nghiệm vì
$ x^4-5x^3+6x^2+5x+1=(x^2-1)^2-5x(x^2-1)+8x^2=(x^2-1-\dfrac{5}{2}x)^2+\dfrac{7}{4}x^2>0 \quad \forall x.$
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