\(\sqrt{2x+\sqrt{4x-1}}+\sqrt{2x-\sqrt{4x-1}}=\sqrt{6}\)
(ĐKXĐ: \(x\ge\dfrac{1}{4}\))
\(\Leftrightarrow2x+\sqrt{4x-1}+2x-\sqrt{4x-1}+2\sqrt{4x^2-4x+1}=6\)
\(\Leftrightarrow4x+2\sqrt{\left(2x-1\right)^2}=6\)
\(\Leftrightarrow2\left(2x+\left|2x-1\right|\right)=6\)
\(\Leftrightarrow2x+\left|2x-1\right|=3\)
\(\Leftrightarrow\left|2x-1\right|=3-2x\)
\(\Leftrightarrow\left[{}\begin{matrix}3-2x\ge0\\2x-1=3-2x\\2x-1=2x-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le\dfrac{3}{2}\\x=1\left(nhận\right)\\0=-2\left(vô.lý\right)\end{matrix}\right.\)
Vậy phương trình đã cho có nghiệm là x=1.
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