Question

\(\sqrt{2x+7}-\sqrt{5-x}\ge\sqrt{3x-2}\)

Answer 1

ĐKXĐ: \(\dfrac{2}{3}\le x\le5\)

\(\sqrt{2x+7}\ge\sqrt{5-x}+\sqrt{3x-2}\)

\(\Leftrightarrow2x+7\ge2x+3+2\sqrt{\left(5-x\right)\left(3x-2\right)}\)

\(\Leftrightarrow\sqrt{-3x^2+17x-10}\le2\)

\(\Leftrightarrow-3x^2+17x-10\le4\)

\(\Leftrightarrow3x^2-17x+14\ge0\Rightarrow\left[{}\begin{matrix}x\ge\dfrac{14}{3}\\x\le1\end{matrix}\right.\)

Kết hợp ĐKXĐ \(\Rightarrow\left[{}\begin{matrix}\dfrac{14}{3}\le x\le5\\\dfrac{2}{3}\le x\le1\end{matrix}\right.\)