Đáp án:
`1)` `B=4/3`
`2)` `A={\sqrt{x}+1}/{\sqrt{x}+2}` với `x\ge 0;x\ne 1`
`3)` `x=0` thì `S_{max}=3/2`
Giải thích các bước giải:
`1)` Khi `x=25`
`=>B={\sqrt{x}+3}/{\sqrt{x}+1}`
`={\sqrt{25}+3}/{\sqrt{25}+1}={5+3}/{5+1}=8/6=4/3`
Vậy `B=4/3` khi `x=25`
`A=\sqrt{x}/{\sqrt{x}-1}+1/{\sqrt{x}+2}-{3\sqrt{x}}/{x+\sqrt{x}-2} `
`\qquad (x\ge 0;x\ne 1)`
`={\sqrt{x}(\sqrt{x}+2)+\sqrt{x}-1-3\sqrt{x}}/{(\sqrt{x}-1)(\sqrt{x}+2)}`
`={x+2\sqrt{x}+\sqrt{x}-1-3\sqrt{x}}/{(\sqrt{x}-1)(\sqrt{x}+2)}`
`={x-1}/{(\sqrt{x}-1)(\sqrt{x}+2)}`
`={(\sqrt{x}-1)(\sqrt{x}+1)}/{(\sqrt{x}-1)(\sqrt{x}+2)}={\sqrt{x}+1}/{\sqrt{x}+2}`
Vậy `A={\sqrt{x}+1}/{\sqrt{x}+2}` với `x\ge 0;x\ne 1`
$\\$
`3)` `S=A.B={\sqrt{x}+1}/{\sqrt{x}+2}.{\sqrt{x}+3}/{\sqrt{x}+1}`
`={\sqrt{x}+3}/{\sqrt{x}+2}={\sqrt{x}+2+1}/{\sqrt{x}+2}`
`={\sqrt{x}+2}/{\sqrt{x}+2}+1/{\sqrt{x}+2}`
`=1+1/{\sqrt{x}+2}`
Với mọi `x\ge 0;x\ne 1` ta có:
`\qquad \sqrt{x}\ge 0`
`=>\sqrt{x}+2\ge 2`
`=>1/{\sqrt{x}+2}\le 1/2`
`=>1+1/{\sqrt{x}+2}\le 1+1/2=3/2`
`=>S\le 3/2`
Dấu "=" xảy ra khi `\sqrt{x}=0<=>x=0\ (thỏa\ mãn)`
Vậy `x=0` thì `S=A.B` có $GTLN$ bằng `3/2`
