⇔x+3>3√x+1⇔{x≥−1(x+3)2>9(x+1)
\Leftrightarrow \begin{cases} x\geq -1\\ x^2-3x>0\end{cases}\Leftrightarrow\left[\begin{array}{I} –1\leq x<0\\ x>3\end{array}\right.
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