\(\dfrac{\left(x-2\right)}{\left(x-3\right)}\ge0\left(\text{đ}k\text{x}\text{đ}\text{ }\ne3\right)\\ \Leftrightarrow x-2\ge0\\ \Leftrightarrow x\ge2\)
\(\dfrac{x-2}{x+3}\ge0\\ \Rightarrow x-2\ge0\\ \Leftrightarrow x\ge2\)
`(x-2)/(x+3)≥0` ĐKXĐ : `x ≠-3`
`=>x-2≥0`
`=>x≥2`
Vậy `S={x|x≥2}`