Question

giải bpt
\(\sqrt{x+2}+\sqrt{3-x}\le x^3+x^2-4x-1\)

Answer 1

ĐKXĐ: \(-2\le x\le3\)

\(\Leftrightarrow3x^3+3x^2-12x-12+x+4-3\sqrt{x+2}+5-x-3\sqrt{3-x}\ge0\)

\(\Leftrightarrow\left(x^2-x-2\right)\left(3x+6\right)+\frac{x^2-x-2}{x+4+3\sqrt{x+2}}+\frac{x^2-x-2}{5-x+3\sqrt{3-x}}\ge0\)

\(\Leftrightarrow\left(x^2-x-2\right)\left[3\left(x+2\right)+\frac{1}{x+4+3\sqrt{x+2}}+\frac{1}{5-x+3\sqrt{3-x}}\right]\ge0\)

\(\Leftrightarrow x^2-x-2\ge0\)

\(\Rightarrow\left[{}\begin{matrix}-2\le x\le-1\\2\le x\le3\end{matrix}\right.\)