Giải :
Ta có :
\(x^4+3x^2-\dfrac{1}{x^4}-\dfrac{3}{x^2-2}=0\)
\(\Leftrightarrow\left(x^4+2x^2+1\right)-\left(\dfrac{1}{x^4}+\dfrac{2}{x^2}+1\right)+x^2-\dfrac{1}{x^2}=0\)\(\Leftrightarrow\left(x^2+1\right)^2-\left(\dfrac{1}{x^2}+1\right)^2+x^2-\dfrac{1}{x^2}=0\)\(\Leftrightarrow\left(x^2+1-\dfrac{1}{x^2}-1\right)\left(x^2+1+\dfrac{1}{x^2}+1\right)+\left(x^2-\dfrac{1}{x^2}\right)=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{x}\right)\left(x+\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}+2\right)+\left(x+\dfrac{1}{x}\right)\left(x-\dfrac{1}{x}\right)=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{x}\right)\left(x+\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}+3\right)=0\)
Vì \(x^2+\dfrac{1}{x^2}+3\ge3\forall x\in R\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{x}=0\\x+\dfrac{1}{x}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{x}\\x=-\dfrac{1}{x}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x^2=1\\x^2=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\VN\end{matrix}\right.\)
Vậy tập nghiệm pt là : \(S=\left\{1;-1\right\}\)
Mình giải sai mất rồi bn ak
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