Question

Giải bất phương trình \(x^2-6x+2\ge2\left(2-x\right)\sqrt{2x-1}\)

Answer 1

Đk: \(x\ge\dfrac{1}{2}\)

Bpt\(\Leftrightarrow\left(x^2+2x\sqrt{2x-1}+2x-1\right)-\left[4\left(2x-1\right)+4\sqrt{2x-1}+1\right]\ge0\)

\(\Leftrightarrow\left(x+\sqrt{2x-1}\right)^2-\left(2\sqrt{2x-1}+1\right)^2\ge0\)

\(\Leftrightarrow\left(x-\sqrt{2x-1}-1\right)\left(x+3\sqrt{2x-1}+1\right)\ge0\) (1)

Vì \(x\ge\dfrac{1}{2}\Rightarrow x+3\sqrt{2x-1}+1>0\)

Từ (1) \(\Rightarrow x-\sqrt{2x-1}-1\ge0\)

\(\Leftrightarrow\sqrt{2x-1}\le x-1\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-1\ge0\\x-1\ge0\\2x-1\le\left(1-x\right)^2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x\in R\backslash\left(2-\sqrt{2};2+\sqrt{2}\right)\end{matrix}\right.\)\(\Rightarrow x\ge2+\sqrt{2}\)

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