\(e,\dfrac{2x-3}{4+x}\ge1\Leftrightarrow\dfrac{2x-3}{4+x}-1\ge0\\ \Leftrightarrow\dfrac{2x-3-4-x}{4+x}\ge0\Leftrightarrow\dfrac{x-7}{4+x}\ge0\\ \Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x\ge-4\end{matrix}\right.\\\left\{{}\begin{matrix}x\le7\\x\le4\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{2x-3}{4+x}< 2\Leftrightarrow\dfrac{2x-3}{4+x}-2< 0\\ \Leftrightarrow\dfrac{2x+3-8-2x}{4+x}< 0\\ \Leftrightarrow\dfrac{-5}{4+x}< 0\Leftrightarrow4+x>0\Leftrightarrow x>\left(-4\right)\)
\(k,\left(2x+1\right)\left(x+3\right)< x^2+3\\ \Leftrightarrow2x^2+7x+3< x^2+3\\ \Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x< 0\\x>\left(-7\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x>0\\x< \left(-7\right)\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow-7< x< 0\rightarrow x\in0\left(ko.có.nghiệm\right)\)
\(-7< 0< 0\)
\(l,\left(4x-1\right)\left(x+3\right)+9>x^2\\ \left\{{}\begin{matrix}\left\{{}\begin{matrix}x>-3\\x>-\dfrac{2}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x< -3\\x< -\dfrac{2}{3}\end{matrix}\right.\end{matrix}\right.\\ x< -3.hoặc.x>-\dfrac{2}{3}\\ \)
Mình sửa câu e nghen, do mình thấy bài bạn Châu làm KTM á
e. \(1\le\dfrac{2x-3}{4+x}\le2\)
\(\left\{{}\begin{matrix}\dfrac{2x-3}{4+x}-1\ge0\\\dfrac{2x-3}{4+x}-2\le0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2x-3-4-x}{4+x}\ge0\\\dfrac{2x-3-8-2x}{4+x}\le0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x-7}{4+x}\ge0\\\dfrac{-11}{4+x}\le0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x-7}{4+x}\ge0\\4+x>0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x-7\ge0\\4+x>0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge7\\x>-4\end{matrix}\right.\)\(\Rightarrow x\ge7\)
l. \(\left(4x-1\right)\left(x+3\right)+9>x^2\)
\(\Leftrightarrow4x^2+11x-3+9>x^2\)
\(\Leftrightarrow3x^2+11x+6>0\)
\(\Leftrightarrow\left[{}\begin{matrix}x< -3\\x>\dfrac{-2}{3}\end{matrix}\right.\)
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