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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/ \(\frac{-25}{\left(-x+2\right)\left(-3x-2\right)}< 0\Leftrightarrow\left[{}\begin{matrix}x< -\frac{2}{3}\\x>2\end{matrix}\right.\)
b/ \(\frac{1}{x-1}-\frac{2}{2x-1}>0\Leftrightarrow\frac{1}{\left(x-1\right)\left(2x-1\right)}>0\Rightarrow\left[{}\begin{matrix}x>1\\x< \frac{1}{2}\end{matrix}\right.\)
c/ \(\frac{2}{3-x}+\frac{2}{x-3}\le0\Leftrightarrow0\le0\) (luôn đúng)
Vậy nghiệm của BPT là \(R\backslash\left\{3\right\}\)
d/ \(1-\frac{x-1}{x^2-3x+2}\ge\Leftrightarrow\frac{x^2-4x+3}{\left(x-1\right)\left(x-2\right)}\ge0\Leftrightarrow\frac{\left(x-1\right)\left(x-3\right)}{\left(x-1\right)\left(x-2\right)}\ge0\)
\(\Rightarrow\left[{}\begin{matrix}x\ge3\\1< x< 2\\x< 1\end{matrix}\right.\)
e/ \(\frac{x+1}{x^2+x+2}-\frac{1}{x+1}>0\Leftrightarrow\frac{x-1}{\left(x+1\right)\left(x^2+x+2\right)}>0\Rightarrow\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\)