Đáp án:
`x=0`
Giải thích các bước giải:
`c)` `A={\sqrt{x}+1}/{2\sqrt{x}+1}`
`\qquad B={\sqrt{x}+2}/{2\sqrt{x}+1}`
`=>P=A+B={\sqrt{x}+1}/{2\sqrt{x}+1}+{\sqrt{x}+2}/{2\sqrt{x}+3}`
`={2\sqrt{x}+1+2}/{2\sqrt{x}+1}`
`={2\sqrt{x}+1}/{2\sqrt{x}+1}+2/{2\sqrt{x}+1}`
`=1+2/{2\sqrt{x}+1}`
Với mọi `x\ge 0;x\ne 1/4`
`=>\sqrt{x}\ge 0`
`=>2\sqrt{x}+1\ge 1`
`=>2/{2\sqrt{x}+1}\le 2`
`=>1+2/{2\sqrt{x}+1}\le 3`
`=>P\le 3`
Vì `2\sqrt{x}+1\ge 1>0`
`=>P=1+2/{2\sqrt{x}+1}>1`
`=>1<P\le 3`
$\\$
Để `P\in NN=>P\in {2;3}`
$\\$
+) `TH: P=2`
`<=>1+2/{2\sqrt{x}+1}=2`
`<=>2/{2\sqrt{x}+1}=1`
`<=>2\sqrt{x}+1=2`
`<=>2\sqrt{x}=1`
`<=>\sqrt{x}=1/2`
`<=>x=1/4\ (loại)`
$\\$
+) `TH: P=3`
`<=>1+2/{2\sqrt{x}+1}=3`
`<=>2/{2\sqrt{x}+1}=2`
`<=>2\sqrt{x}+1=1`
`<=>2\sqrt{x}=0`
`<=>x=0\ (thỏa\ đk)`
Vậy `x=0` thỏa đề bài
