\(BPT\Leftrightarrow\dfrac{\left(x-2\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\ge\dfrac{\left(3x+2\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{2\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(\Rightarrow x^2-x-2x+2-3x^2-3x-2x-2-2x^2-2\ge0\)
\(\Leftrightarrow-4x^2-8x-2\ge0\)
\(\Leftrightarrow x^2+2x+\dfrac{1}{2}\ge0\)
\(\Leftrightarrow\left(x+1\right)^2-\dfrac{1}{2}\ge0\)
Vậy bất phương trình luôn đúng \(\forall x\).
ĐKXĐ: \(x\ne1,-1\)
Ta có: \(\dfrac{x-2}{x+1}\ge\dfrac{3x+2}{x-1}-2\)
\(\dfrac{x-2}{x+1}\ge\dfrac{3x+2-2\left(x-1\right)}{x-1}\)
\(\dfrac{x-2}{x+1}-\dfrac{3x+2-2x+2}{x-1}\ge0\)
\(\dfrac{x-2}{x+1}-\dfrac{x+4}{x-1}\ge0\)
\(\dfrac{\left(x-2\right)\left(x-1\right)-\left(x-4\right)\left(x+1\right)}{x^2-1}\ge0\)
\(\dfrac{x^2-3x+2-x^2+3x+4}{x^2-1}\ge0\)
\(\dfrac{6}{x^2-1}\ge0\)
\(\Rightarrow x^2-1>0\Leftrightarrow x^2>1\Leftrightarrow\left\{{}\begin{matrix}x< -1\\x>1\end{matrix}\right.\)(TM)
Biểu thức vế trái có nghĩa khi
\(x\ne-2;x\ne1\\ \dfrac{x-2}{x+2}+\dfrac{x+1}{x-1}\Leftrightarrow\dfrac{x-2}{x+2}+\dfrac{x+1}{x-1}-2>0\\ \dfrac{\left(x-2\right)\left(x-1\right)+\left(x+2\right)\left(x+1\right)-2\left(x+2\right)\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}\\ >0\\ \Leftrightarrow\dfrac{8-2x}{\left(x+2\right)\left(x-1\right)}>0\\ \Leftrightarrow\dfrac{4-x}{\left(x+2\right)\left(x-1\right)}>0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)\left(x-4\right)< 0\)
Lập bảng xét dấu
| x | -2 | 1 | 4 |
| x+2 | - 0 + | + | + |
| x-1 | - - | 0 + | + |
| x-4 | - - | - | 0 + |
| VT | - 0 + | 0 - | 0 + |
Vậy nghiệm của bất pt là
\(x< -2.hay.1< x< 4\)
