Đáp án:
`6)A=(6sqrtx+8)/(3sqrtx+2)`
`A=(6sqrtx+4+4)/(3sqrtx+2)`
`A=(2(3sqrtx+2)+4)/(3sqrtx+2)`
`A=2+4/(3sqrtx+2)`
Vì `4/(3sqrtx+2)>2AAx>=0`
`=>A>2+0=2(1)`
Vì `3sqrtx+2>=2>0`
`<=>4/(3sqrtx+2)<=4/2=2`
`<=>A<=2+2=4(2)`
`(1)(2)=>2<A<=4`
Mà `A in ZZ`
`=>A in {3,4}`
`**A=3`
`<=>2+4/(3sqrtx+2)=3`
`<=>4/(3sqrtx+2)=1`
`<=>3sqrtx+2=4`
`<=>3sqrtx=2`
`<=>sqrtx=2/3`
`<=>x=4/9`
`**A=4`
`<=>2+4/(3sqrtx+2)=4`
`<=>4/(3sqrtx+2)=2`
`<=>6sqrtx+4=4`
`<=>6sqrtx=0`
`<=>x=0`.
Vậy với `x in {0,4/9}` thì `A in ZZ`.
