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a/
\(\Leftrightarrow\frac{\left(x^2-1\right)\left(x^2+1\right)}{x^2+3x}+x^2-1\ge0\)
\(\Leftrightarrow\left(x^2-1\right)\left(\frac{x^2+1}{x^2+3x}+1\right)\ge0\)
\(\Leftrightarrow\left(x^2-1\right)\left(\frac{2x^2+3x+1}{x^2+3x}\right)\ge0\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x+1\right)\left(x+1\right)\left(2x+1\right)}{x\left(x+3\right)}\ge0\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(2x+1\right)\left(x+1\right)^2}{x\left(x+3\right)}\ge0\)
\(\Rightarrow\left[{}\begin{matrix}x< -3\\x=-1\\-\frac{1}{2}\le x< 0\\x\ge1\end{matrix}\right.\)
b/
\(\Leftrightarrow\left(x^2-1\right)\left(x^2-4\right)\left(\frac{-2-2x}{x}\right)\le0\)
\(\Leftrightarrow\frac{-2.\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\left(x+1\right)}{x}\le0\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(x-1\right)\left(x-2\right)\left(x+1\right)^2}{x}\ge0\)
\(\Rightarrow\left[{}\begin{matrix}x\le-2\\x=-1\\0< x\le1\\x\ge2\end{matrix}\right.\)
c/
\(\Leftrightarrow\left(\frac{4\left(x-1\right)-2x}{x\left(x-1\right)}\right)\left(\frac{x^2+1-2x}{x}\right)\le0\)
\(\Leftrightarrow\frac{\left(2x-4\right)\left(x-1\right)^2}{x^2\left(x-1\right)}\le0\)
\(\Leftrightarrow\frac{\left(x-2\right)\left(x-1\right)^2}{x^2\left(x-1\right)}\le0\)
\(\Rightarrow1< x\le2\)
a/ \(\Leftrightarrow\frac{2x\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{x\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}\le0\)
\(\Leftrightarrow\frac{5\left(-3x+4\right)}{\left(x-5\right)\left(x+5\right)}\le0\) \(\Rightarrow\left[{}\begin{matrix}-5< x\le\frac{4}{3}\\x>5\end{matrix}\right.\)
b/ Không rõ đề
c/
- Với \(x< -1\Rightarrow\left\{{}\begin{matrix}VT>0\\VP< 0\end{matrix}\right.\) BPT vô nghiệm
- Với \(x\ge-1\) hai vế ko âm, bình phương:
\(\Leftrightarrow\left(x+1\right)^2>\frac{\left(x-3\right)^2}{4}\)
\(\Leftrightarrow\left(2x+2\right)^2-\left(x-3\right)^2>0\)
\(\Leftrightarrow\left(x+5\right)\left(3x-1\right)>0\Rightarrow\left[{}\begin{matrix}x< -5\\x>\frac{1}{3}\end{matrix}\right.\) \(\Rightarrow x>\frac{1}{3}\)
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. \(\Leftrightarrow\frac{4}{\left(x-2\right)^2}+\frac{1}{x-2}>0\Leftrightarrow\frac{x+2}{\left(x-2\right)^2}\ge0\Rightarrow\left[{}\begin{matrix}x\ge-2\\x\ne2\end{matrix}\right.\)