1.
\(\Leftrightarrow\left\{{}\begin{matrix}2x+1\ge0\\\left(2x+1\right)^2\ge x^2+3x+5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{1}{2}\\4x^2+4x+1\ge x^2+3x+5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{1}{2}\\3x^2+x-4\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{1}{2}\\\left[{}\begin{matrix}x\le-\dfrac{4}{3}\\x\ge1\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow x\ge1\)
2.
\(\Leftrightarrow\left\{{}\begin{matrix}x-3>0\\x^2+x-12\ge0\\\left(x-3\right)^2>x^2+x-12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>3\\x^2+x-12\ge0\\7x< 21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>3\\\left[{}\begin{matrix}x\le-4\\x\ge3\end{matrix}\right.\\x< 3\end{matrix}\right.\)
\(\Rightarrow x\in\varnothing\)
BPT đã cho vô nghiệm
3.
\(\Leftrightarrow x-1>\sqrt{2x^2-3x-5}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1>0\\2x^2-3x-5\ge0\\\left(x-1\right)^2>2x^2-3x-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>1\\2x^2-3x-5\ge0\\x^2-x-6< 0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>1\\\left[{}\begin{matrix}x\le-1\\x\ge\dfrac{5}{3}\end{matrix}\right.\\-2< x< 3\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{5}{3}\le x< 3\)
4.
\(\Leftrightarrow\left\{{}\begin{matrix}5x+1>0\\\left(5x+1\right)^2>4\left(x^2+2x+6\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>-\dfrac{1}{5}\\21x^2+2x-23>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>-\dfrac{1}{5}\\\left[{}\begin{matrix}x>1\\x< -\dfrac{23}{21}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow x>1\)
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