a/ \(\dfrac{x-1}{x+1}-\dfrac{x}{x+2}=\dfrac{x-3}{\left(x+1\right)\left(x+2\right)}\) ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\x\ne-2\end{matrix}\right.\)
\(\Rightarrow\left(x-1\right)\left(x+2\right)-x\left(x+1\right)=x-3\)
\(\Leftrightarrow x^2+2x-x-2-x^2-x=x-3\)
\(\Leftrightarrow-x=-1\Leftrightarrow x=1\left(tm\right)\)
Vậy...............................
b/ \(\dfrac{x-3}{2}\ge0\Leftrightarrow x-3\ge0\Leftrightarrow x\ge3\)
Vậy....................
c/ ĐK: x khác 2
\(\dfrac{x+1}{x-2}>0\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1>0\\x-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1< 0\\x-2< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-1\\x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x< -1\\x< 2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -1\end{matrix}\right.\left(tm\right)\)
Vậy.................
d/ \(x^2-x-6\ge0\)
\(\Leftrightarrow x^2-3x+2x-6\ge0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)\ge0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3\ge0\\x+2\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3\le0\\x+2\le0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge3\\x\ge-2\end{matrix}\right.\\\left\{{}\begin{matrix}x\le3\\x\le-2\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\ge3\\x\le-2\end{matrix}\right.\)
Vậy x ≥ 3 hoặc x ≤-2
