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$x^4-2x^2+\dfrac{1}{2}=0\Leftrightarrow x^4-2x^2+1=\dfrac{1}{2}\Leftrightarrow (x^2-1)^2=\dfrac{1}{2}$
$\Leftrightarrow x^2-1 =\pm\dfrac{1}{\sqrt 2}\Leftrightarrow x^2=1\pm\dfrac{1}{\sqrt 2}\Leftrightarrow x=\pm\sqrt{1\pm\dfrac{1}{\sqrt 2}}$.
Vậy Pt này có bốn nghiệm
$x \in \left\{ {\sqrt{1+\dfrac{1}{\sqrt 2}},\sqrt{1-\dfrac{1}{\sqrt 2}},-\sqrt{1+\dfrac{1}{\sqrt 2}},-\sqrt{1-\dfrac{1}{\sqrt 2}}} \right\}$.
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