`(x-2)(-x^2-5x+6)>=0`
`<=>(x-2)(x^2+5x-6)<=0`
`<=>`$\left[ \begin{array}{l}\begin{cases}x-2 \geq 0\\x^2+5x-6 \leq 0\end{cases}\\\begin{cases}x-2 \leq 0\\x^2+5x-6 \geq 0\end{cases}\end{array} \right.$
`<=>`$\left[ \begin{array}{l}\begin{cases}x \ge q 2\\(x-1)(x+6) \leq 0 end{cases}\\\begin{cases}x \leq 2\\(x-1)(x+6) \geq 0\end{cases}\end{array} \right.$
`<=>`$\left[ \begin{array}{l}\begin{cases}x \geq 2\\-6 \leq x \leq 1\end{cases}(\text{vô lý})\\\begin{cases}x \leq 2\\\left[ \begin{array}{l}x \ge 1\\x \leq -6\end{array} \right.\end{cases}\end{array} \right.$
`<=>` \left[ \begin{array}{l}1 \leq x \leq 2\\x \leq -6\end{array} \right.
Vậy............
`(x-2)(-x^2-5x+6)>=0`
`<=>(x-2)(x^2+5x-6)<=0`
`<=>`$\left[ \begin{array}{l}\begin{cases}x-2 \geq 0\\x^2+5x-6 \leq 0\end{cases}\\\begin{cases}x-2 \leq 0\\x^2+5x-6 \geq 0\end{cases}\end{array} \right.$
`<=>`$\left[ \begin{array}{l}\begin{cases}x \ge q 2\\(x-1)(x+6) \leq 0 end{cases}\\\begin{cases}x \leq 2\\(x-1)(x+6) \geq 0\end{cases}\end{array} \right.$
`<=>`$\left[ \begin{array}{l}\begin{cases}x \geq 2\\-6 \leq x \leq 1\end{cases}(\text{vô lý})\\\begin{cases}x \leq 2\\\left[ \begin{array}{l}x \geq 1\\x \leq -6\end{array} \right.\end{cases}\end{array} \right.$
`<=>` $\left[ \begin{array}{l}1 \leq x \leq 2\\x \leq -6\end{array} \right.$
`(x-2)(-x^2-5x+6)>=0`
`<=>(x-2)(x^2+5x-6)<=0`
`<=>`$\left[ \begin{array}{l}\begin{cases}x-2 \geq 0\\x^2+5x-6 \leq 0\end{cases}\\\begin{cases}x-2 \leq 0\\x^2+5x-6 \geq 0\end{cases}\end{array} \right.$
`<=>`$\left[ \begin{array}{l}\begin{cases}x \geq 2\\(x-1)(x+6) \leq 0 end{cases}\\\begin{cases}x \leq 2\\(x-1)(x+6) \geq 0\end{cases}\end{array} \right.$
`<=>`$\left[ \begin{array}{l}\begin{cases}x \geq 2\\-6 \leq x \leq 1\end{cases}(\text{vô lý})\\\begin{cases}x \leq 2\\\left[ \begin{array}{l}x \geq 1\\x \leq -6\end{array} \right.\end{cases}\end{array} \right.$
`<=>` $\left[ \begin{array}{l}1 \leq x \leq 2\\x \leq -6\end{array} \right.$
Phù ==""
