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1. \(\Leftrightarrow\left(3x-1\right)\left(\sqrt{5}x-2\right)\ge0\Rightarrow\left[{}\begin{matrix}x\le\frac{1}{3}\\x\ge\frac{2}{\sqrt{5}}\end{matrix}\right.\)
2. \(\Leftrightarrow\frac{\left(3-2x\right)\left(3+2x\right)}{2x-3}\ge0\Leftrightarrow\left[{}\begin{matrix}x\ne\frac{3}{2}\\x\le-\frac{3}{2}\end{matrix}\right.\)
3. \(\left|x-2\right|\ge3\Leftrightarrow\left[{}\begin{matrix}x-2\ge3\\x-2\le-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x\ge5\\x\le-1\end{matrix}\right.\)
4. \(\Leftrightarrow-10\le3x+1\le10\Rightarrow-\frac{11}{3}\le x\le3\)
5. \(\Leftrightarrow\frac{3x^2-x+2}{x^2-9}-3\le0\Leftrightarrow\frac{-x+29}{\left(x-3\right)\left(x+3\right)}\le0\Rightarrow\left[{}\begin{matrix}-3< x< 3\\x\ge29\end{matrix}\right.\)
6. \(\Leftrightarrow\frac{4}{\left(x-2\right)^2}+\frac{1}{x-2}>0\Leftrightarrow\frac{x+2}{\left(x-2\right)^2}\ge0\Rightarrow\left[{}\begin{matrix}x\ge-2\\x\ne2\end{matrix}\right.\)
a, ĐKXĐ : \(x^2+2\ne0\) ( luôn đúng với mọi x )
Ta có :\(\frac{3x-2}{x^2+2}\ge-1\)
=> \(\frac{3x-2}{x^2+2}+1\ge0\)
=> \(\frac{3x-2+x^2+2}{x^2+2}\ge0\)
=> \(x^2+3x\ge0\)
=> \(x\left(x+3\right)\ge0\)
=> \(\left[{}\begin{matrix}x\ge0\\x+3\ge0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x\ge0\\x\ge-3\end{matrix}\right.\)
b, Ta có : \(\left|5x-9\right|\le6\)
TH1 : \(5x-9\ge0\left(x\ge\frac{9}{5}\right)\)
=> \(5x-9\le6\)
=> \(5x\le15\)
=> \(x\le3\)
=> \(\frac{9}{5}\le x\le3\)
TH2 : \(5x-9< 0\left(x< \frac{9}{5}\right)\)
=> \(9-5x\le6\)
=> \(-5x\le-3\)
=> \(x\ge\frac{3}{5}\)
=> \(\frac{3}{5}\le x< \frac{9}{5}\)