Lời giải.
ĐKXĐ: `x≥0.x\ne1.`
Ta có: `B={3\sqrt{x}+1}/{x+2\sqrt{x}-3}-2/{\sqrt{x}+3}`
`B={3\sqrt{x}+1}/{(\sqrt{x}+3)(\sqrt{x}-1)}-{2(\sqrt{x}-1)}/{(\sqrt{x}+3)(\sqrt{x}-1)}`
`B={3\sqrt{x}+1-2\sqrt{x}+2}/{(\sqrt{x}+3)(\sqrt{x}-1)}`
`B={\sqrt{x}+3}/{(\sqrt{x}+3)(\sqrt{x}-1)}=1/{\sqrt{x}-1}.`
Ta có: `A/B=A:B={\sqrt{x}+4}/{\sqrt{x}-1}:1/{\sqrt{x}-1}={\sqrt{x}+4}/{\sqrt{x}-1}.(\sqrt{x}-1)`
`=\sqrt{x}+4.`
Có: `\sqrt{x}≥0∀x=>\sqrt{x}+4≥4<=>A/B≥4.`
Dấu "=" xảy ra khi `x=0.`
Vậy ta có điều phải chứng minh.
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Lưu ý: `x+2\sqrt{x}-3=x+3\sqrt{x}-\sqrt{x}-3=\sqrt{x}(\sqrt{x}+3)-(\sqrt{x}+3)=(\sqrt{x}+3)(\sqrt{x}-1).`
