Đáp án:
Điều kiện:`(x^2-3x+8)/(x-3)>=0(x ne 3)`
Mà `x^2-3x+8`
`=(x-3/2)^2+23/4>=23/4>0`
`<=>x-3>0`
`<=>x>3.`
`pt<=>\sqrt{x^2-3x+8}=3sqrt{x-3}`
`<=>x^2-3x+8=9(x-3)`
`<=>x^2-12x+35=0`
`Delta'=36-35=1`
`<=>` \(\left[ \begin{array}{l}x_1=\dfrac{-b'+\sqrt{\Delta'}}{a}=7\\x_2=\dfrac{-b'-\sqrt{\Delta'}}{a}=5(tm)\end{array} \right.\)
Vậy `S={5,7}.`