Đáp án:
Giải thích các bước giải:
`A=(4\sqrt{x}+9)/(2\sqrt{x}+1)`
`ĐK:x>=0`
Ta có `2\sqrt{x}+1>=1`
`=>7/(2\sqrt{x}+1)<=7`
`=>A<=9`
Lại có `A=2+7/(2\sqrt{x}+1)>2`
`=>2<A<=9`
Mà `A in Z=>A in {3,4,5,6,7,8,9}`
`=>7/(2\sqrt{x}+1) in {1,2,3,4,5,6,7}`
$\bullet$
`7/(2\sqrt{x}+1) =1`
`=>2\sqrt{x}+1=7`
`=>2\sqrt{x}=6`
`=>\sqrt{x}=3`
`=>x=9`
$\bullet$
`7/(2\sqrt{x}+1) =2`
`=>2\sqrt{x}+1=7/2`
`=>2\sqrt{x}=5/2`
`=>\sqrt{x}=5/4`
`=>x=25/16`
$\bullet$
`7/(2\sqrt{x}+1) =3`
`=>2\sqrt{x}+1=7/3`
`=>2\sqrt{x}=4/3`
`=>\sqrt{x}=2/3`
`=>x=4/9`
$\bullet$
`7/(2\sqrt{x}+1) =4`
`=>2\sqrt{x}+1=7/4`
`=>2\sqrt{x}=3/4`
`=>\sqrt{x}=3/8`
`=>x=9/64`
$\bullet$
`7/(2\sqrt{x}+1) =5`
`=>2\sqrt{x}+1=7/5`
`=>2\sqrt{x}=2/5`
`=>\sqrt{x}=1/5`
`=>x=1/25`
$\bullet$
`7/(2\sqrt{x}+1) =6`
`=>2\sqrt{x}+1=7/6`
`=>2\sqrt{x}=1/6`
`=>\sqrt{x}=1/12`
`=>x=1/144`
$\bullet$
`7/(2\sqrt{x}+1) =7`
`=>2\sqrt{x}+1=1`
`=>2\sqrt{x}=0`
`=>\sqrt{x}=0`
`=>x=0`
