1) \(\dfrac{5}{x-3}+\dfrac{3}{x+2}\le\dfrac{3+2x}{x^2-x-6}\)2) \(\dfrac{1}{x^2-4}+\dfrac{2}{x+2}< \dfrac{-3}{x-2}\)3) (4-x-\(3x^2\)).(x+2).(x+1) > 04) (\(x^3\)-9x).(x-3) ≥ 05) \(\left|4-x\right|\) ≥ 2x-16) \(\left|x-2\right|\) ≤...
Đọc tiếp
1) \(\dfrac{5}{x-3}+\dfrac{3}{x+2}\le\dfrac{3+2x}{x^2-x-6}\)
2) \(\dfrac{1}{x^2-4}+\dfrac{2}{x+2}< \dfrac{-3}{x-2}\)
3) (4-x-\(3x^2\)).(x+2).(x+1) > 0
4) (\(x^3\)-9x).(x-3) ≥ 0
5) \(\left|4-x\right|\) ≥ 2x-1
6) \(\left|x-2\right|\) ≤ 1-x
7) \(\left|x+2\right|+2x-3\le0\)
8) \(\sqrt{x^2+6x+9}-2x+1>0\)
1.
ĐK: \(x\ne3;x\ne-2\)
\(\dfrac{5}{x-3}+\dfrac{3}{x+2}\le\dfrac{3+2x}{x^2-x-6}\)
\(\Leftrightarrow\dfrac{5\left(x+2\right)+3\left(x-3\right)}{x^2-x-6}\le\dfrac{3+2x}{x^2-x-6}\)
\(\Leftrightarrow\dfrac{8x+1-3-2x}{x^2-x-6}\le0\)
\(\Leftrightarrow\dfrac{6x-2}{x^2-x-6}\le0\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x-2\ge0\\x^2-x-6< 0\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}6x-2\le0\\x^2-x-6>0\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}6x-2\ge0\\x^2-x-6< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{3}\\-2< x< 3\end{matrix}\right.\Leftrightarrow\dfrac{1}{3}\le x< 3\)
TH2: \(\left\{{}\begin{matrix}6x-2\le0\\x^2-x-6>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{3}\\\left[{}\begin{matrix}x>3\\x< -2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow x< -2\)
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2.
ĐK: \(x\ne\pm2\)
\(\dfrac{1}{x^2-4}+\dfrac{2}{x+2}>-\dfrac{3}{x-2}\)
\(\Leftrightarrow\dfrac{1}{x^2-4}+\dfrac{2\left(x-2\right)+3\left(x+2\right)}{x^2-4}>0\)
\(\Leftrightarrow\dfrac{5x+3}{x^2-4}>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}5x+3>0\\x^2-4>0\end{matrix}\right.\\\left\{{}\begin{matrix}5x+3< 0\\x^2-4< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-\dfrac{3}{5}< x< 2\\x< -2\end{matrix}\right.\)
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