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ĐK: $x \geqslant - 1$ $\begin{array} PT \Leftrightarrow {(2x)^3} + 2x < (x + 1 + 1)\sqrt {x + 1} \\ \,\,\,\,\,\,\, \Leftrightarrow {(2x)^3} + 2x < {(\sqrt {x + 1} )^3} + \sqrt {x + 1} \\ \,\,\,\,\,\,\, \Leftrightarrow f(2x) < f(\sqrt {x + 1} ),\,\,\,f(t) = {t^3} + t \\ \,\,\,\,\,\,\, \Leftrightarrow 2x < \sqrt {x + 1} \\ \,\,\,\,\,\,\, \Leftrightarrow \left[ \begin{array} x < 0 \\ \left\{ \begin{array} x \geqslant 0 \\ 4{x^2} < x + 1 \\ \end{array} \right. \\ \end{array} \right. \Leftrightarrow \left[ \begin{array} x < 0 \\ \left\{ \begin{array} x \geqslant 0 \\ 0 < x < \frac{{1 + \sqrt {17} }}{8} \\ \end{array} \right. \\ \end{array} \right. \\ \end{array} $ Vậy bất phương trình có nghiệm $ - 1 \leqslant x < \frac{{1 + \sqrt {17} }}{8}$
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