Question

\(\sqrt{2x^2+3x+1}+\sqrt{1-3x}=2\sqrt{2x^2+1}\)

Answer 1

ĐK:\(-1\le x\le\dfrac{1}{3}\)

Ta có: VT=\(\sqrt{2x^2+3x+1}+\sqrt{1-3x}\le\sqrt{\left(1+1\right)\left(\sqrt{2x^2+3x+1}^2+\sqrt{1-3x}^2\right)}\)

\(=\sqrt{2.\left(2x^2+2\right)}=2\sqrt{x^2+1}\)

Xét VT= \(2\sqrt{2x^2+1}\ge2\sqrt{x^2+1}\)

\(\Leftrightarrow2x^2+1\ge x^2+1\Leftrightarrow x^2\ge0\) (đúng)

\(\Rightarrow VP\ge VT\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{\sqrt{2x^2+3x+1}}=\dfrac{1}{\sqrt{1-3x}}\\2\sqrt{2x^2+1}=2\sqrt{x^2+1}\end{matrix}\right.\Leftrightarrow x=0\left(tm\right)\)