Đáp án:
`a)``{2sqrtx-9}/{(sqrtx-2)(sqrtx-3)}-{sqrtx+3}/{sqrtx-2}-{2sqrtx+1}/{3-sqrtx)`
`={2sqrtx-9-(sqrtx+3)(sqrtx-3)}/{(sqrtx-2)(sqrtx-3)}+{2sqrtx+1}/(sqrtx-3)`
`={2sqrtx-9-(x-9)+(2sqrtx+1)(sqrtx-2)}/{(sqrtx-2)(sqrtx-3)}`
`={2sqrtx-9-x+9+2x-4sqrtx+sqrtx-2}/{(sqrtx-2)(sqrtx-3)}`
`={x-sqrtx-2}/{(sqrtx-2)(sqrtx-3)}={x-2sqrtx+sqrtx-2}/{(sqrtx-2)(sqrtx-3)}={sqrtx(sqrtx-2)+(sqrtx-2)}/{(sqrtx-2)(sqrtx-3)}={(sqrtx+1)(sqrtx-2)}/{(sqrtx-2)(sqrtx-3)}={sqrtx+1}/{sqrtx-3}`
`b)``Q=2=>{sqrtx+1}/{sqrtx-3}=2<=>sqrtx+1=2sqrtx-6`
`<=>sqrtx=7`
`<=>x=49`
`c)` Ta có: `Q={sqrtx+1}/{sqrtx-3}={sqrtx-3+4}/{sqrtx-3}=1+4/{sqrtx-3}`
Để `Q` nguyên `<=>4vdots(sqrtx-3)<=>(sqrtx-3)` là ước của `4`
`=>`\(\left[ \begin{array}{l}\sqrt{x}-3=1\\\sqrt{x}-3=-1\\\sqrt{x}-3=2\\\sqrt{x}-3=-2\\\sqrt{x}-3=4\\\sqrt{x}-3=-4\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}\sqrt{x}=4⇒x=16\\\sqrt{x}=2⇒x=4 (\text{loại vì} x\ne4)\\\sqrt{x}=5⇒x=25\\\sqrt{x}=1⇒x=1\\\sqrt{x}=7⇒x=49\\\sqrt{x}=-1(\text{vô lí vì }\sqrt{x}\ge0)\end{array} \right.\)
Vậy để `Q` nguyên thì `x in{16,25,1,49}`
