ĐKXĐ: \(x\ge0\)
\(\Leftrightarrow x^2+4+2x=3\sqrt{x\left(x^2+4\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\ge0\\\sqrt{x^2+4}=b>0\end{matrix}\right.\)
\(\Rightarrow b^2+2a^2=3ab\)
\(\Leftrightarrow2a^2-3ab+b^2=0\)
\(\Leftrightarrow\left(a-b\right)\left(2a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\b=2a\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\sqrt{x^2+4}\\\sqrt{x^2+4}=2\sqrt{x}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+4=0\left(vn\right)\\x^2-4x+4=0\end{matrix}\right.\) \(\Rightarrow x=2\)
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