ĐK: \(x\ge-\dfrac{3}{2}\)
\(\sqrt{2x+3}+\sqrt{x+2}\le1\)
\(\Leftrightarrow3x+5+2\sqrt{2x^2+7x+6}\le1\)
\(\Leftrightarrow2\sqrt{2x^2+7x+6}\le-3x-4\)
\(\Leftrightarrow\left\{{}\begin{matrix}-3x-4\ge0\\4\left(2x^2+7x+6\right)\le\left(3x+4\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le-\dfrac{4}{3}\\8x^2+28x+24\le9x^2+16+24x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le-\dfrac{4}{3}\\x^2-4x-8\ge0\end{matrix}\right.\)
\(\Leftrightarrow x\le2-2\sqrt{3}\)
Vậy \(-\dfrac{3}{2}\le x\le2-2\sqrt{3}\)
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