\(x-\frac{11x^2-5x+6}{x^2+5x+6}>0\)
\(\Leftrightarrow\frac{x^3-6x^2+11x-6}{x^2+5x+6}>0\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x-2\right)\left(x-3\right)}{\left(x+2\right)\left(x+3\right)}>0\Rightarrow\left[{}\begin{matrix}x>3\\1< x< 2\\-3< x< -2\end{matrix}\right.\)
b/ \(\frac{2-x}{x^3+x^2}-\frac{1-2x}{x^3-3x^2}>0\)
\(\Leftrightarrow\frac{\left(2-x\right)\left(x+1\right)-\left(1-2x\right)\left(x-3\right)}{x^2\left(x+1\right)\left(x-3\right)}>0\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x-5\right)}{x^2\left(x+1\right)\left(x-3\right)}>0\Rightarrow\left[{}\begin{matrix}x< -1\\x>5\\1< x< 3\end{matrix}\right.\)
c/ \(\left|x^2-x-1\right|\le x-1\)
Với \(x< 1\Rightarrow\left\{{}\begin{matrix}VT\ge0\\VP< 0\end{matrix}\right.\) BPT vô nghiệm
Với \(x\ge1\) hai vế ko âm, bình phương:
\(\left(x^2-x-1\right)^2\le\left(x-1\right)^2\)
\(\Leftrightarrow\left(x^2-x-1\right)^2-\left(x-1\right)^2\le0\)
\(\Leftrightarrow\left(x^2-2x\right)\left(x^2-2\right)\le0\) \(\Rightarrow\sqrt{2}\le x\le2\)
