$\color{#cff0f3}{>} \color{#c9ecf9}{<} \color{#d2e4f7}{h} \color{#d2e0f7}{h} \color{#c1d8fc}{2} \color{#aac4ee}{0} \color{#8cb2f0}{0} \color{#3d85c6}{5}$

`1.`

`a)`

Tại `x=4`

`A=(2\sqrt4-3)/(\sqrt4+1)=(2*2-3)/(2+1)=(4-3)/3=1/3`

`B=(6\sqrt4+1)/(3\sqrt4-2)=(6*2+1)/(3*2-2)=(12+1)/(6-2)=13/4`

Tại `x=16`

`A=(2\sqrt16-3)/(\sqrt16+1)=(2*4-3)/(4+1)=(8-3)/5=5/5=1`

`B=(6\sqrt16+1)/(3\sqrt16-2)=(6*4+1)(3*4-2)=(24+1)/(12-2)=25/10=5/2`

`b)`

Để `A>0`

`=>(2\sqrtx-3)/(\sqrtx+1)>0`

`=>2\sqrtx-3>0  (\sqrtx+1>0)`

`=>2\sqrtx>3`

`=>\sqrtx>3/2`

`=>x>9/4`

Để `B>0`

`=>(6\sqrtx+1)/(3\sqrtx-2)>0`

`=>3\sqrtx-2>0  (6\sqrtx+1>0`

`=>3\sqrtx>2`

`=>\sqrtx>2/3`

`=>x>4/9`

`c)`

Để `A>=0`

`=>(2\sqrtx-3)/(\sqrtx+1)>=0`

`=>2\sqrtx-3>=0  (\sqrtx+1>0)`

`=>2\sqrtx>=3`

`=>\sqrtx>=3/2`

`=>x>=9/4`

Để `B>=0`

`=>(6\sqrtx+1)/(3\sqrtx-2)>=0`

`=>3\sqrtx-2>0  (6\sqrtx+1>0)`

`=>3\sqrtx>2`

`=>\sqrtx>2/3`

`=>x>4/9`

`d)`

Để `A in ZZ`

`=>(2\sqrtx-3)/(\sqrtx+1) in ZZ`

`=>2\sqrtx-3 vdots \sqrtx+1`

`=>2(\sqrtx+1)-(2\sqrtx-3) vdots \sqrtx+1`

`=>5 vdots \sqrtx+1`

`=>\sqrtx+1 in Ư_((5))={1;5}`

`=>x in {0;8}`

Để `B in ZZ`

`=>(6\sqrtx+1)/(3\sqrtx-2) in ZZ`

`=>6\sqrtx+1 vdots 3\sqrtx-2`

`=>(6\sqrtx+1)-2(3\sqrtx-2) vdots 3\sqrtx-2`

`=>5 vdots 3\sqrtx-2`

`=>3\sqrtx-2 in Ư_((5))={+-1;+-5}`

`=>x in {1/9;1;64/9}`

mà `x in ZZ`

`=>x in {1}`

Vậy `...`