Question

\(\sqrt{2x+4}=\frac{6x-4}{\sqrt{x^2+4}}+2\sqrt{2-x}\)

Answer 1

ĐKXĐ: ...

\(\Leftrightarrow\sqrt{2x+4}-2\sqrt{2-x}-\frac{6x-4}{\sqrt{x^2+4}}=0\)

\(\Leftrightarrow\frac{6x-4}{\sqrt{2x+4}+2\sqrt{2-x}}-\frac{6x-4}{\sqrt{x^2+4}}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}6x-4=0\\\sqrt{2x+4}+2\sqrt{2-x}=\sqrt{x^2+4}\left(1\right)\end{matrix}\right.\)

Xét (1):

\(VT=\sqrt{2x+4}+\sqrt{8-4x}\ge\sqrt{12-2x}\ge\sqrt{12-2.2}=2\sqrt{2}\)

\(VP=\sqrt{x^2+4}\le\sqrt{2^2+4}=2\sqrt{2}\)

\(\Rightarrow VT\ge VP\)

Dấu "=" xảy ra khi và chỉ khi \(x=2\)