Question

giải bất phương trình :

\(\sqrt{x+2}+\sqrt{4-x}>\sqrt{2x+3}\)

Answer 1

ĐKXĐ: \(-\dfrac{3}{2}\le x\le4\)

BPT tương đương:

\(6+2\sqrt{\left(x+2\right)\left(4-x\right)}>2x+3\)

\(\Leftrightarrow2\sqrt{-x^2+2x+8}>2x-3\)

\(\Leftrightarrow\left[{}\begin{matrix}x< \dfrac{3}{2}\\\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\4\left(-x^2+2x+8\right)>4x^2-12x+9\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x< \dfrac{3}{2}\\\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\8x^2-20x-23< 0\end{matrix}\right.\end{matrix}\right.\) 

\(\Rightarrow-\dfrac{3}{2}\le x< \dfrac{5+\sqrt{71}}{4}\)