`a)x=16` (t/m đk)
`=>A=[16+7]/\sqrt{16}=23/4`
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`b)` Với `x > 0,x \ne 9` có:
`B=\sqrt{x}/[\sqrt{x}+3]+[2\sqrt{x}-1]/[\sqrt{x}-3]-[2x-\sqrt{x}-3]/[x-9]`
`B=[\sqrt{x}(\sqrt{x}-3)+(2\sqrt{x}-1)(\sqrt{x}+3)-2x+\sqrt{x}+3]/[(\sqrt{x}-3)(\sqrt{x}+3)]`
`B=[x-3\sqrt{x}+2x+6\sqrt{x}-\sqrt{x}-3-2x+\sqrt{x}+3]/[(\sqrt{x}-3)(\sqrt{x}+3)]`
`B=[x+3\sqrt{x}]/[(\sqrt{x}-3)(\sqrt{x}+3)]`
`B=\sqrt{x}/[\sqrt{x}-3]`
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`c)` Với `x > 0,x \ne 9` có:
`S=1/B+A=1:\sqrt{x}/[\sqrt{x}-3]+[x+7]/\sqrt{x}`
`=[\sqrt{x}-3]/\sqrt{x}+[x+7]/\sqrt{x}`
`=[x+\sqrt{x}+4]/\sqrt{x}=\sqrt{x}+1+4/\sqrt{x}`
Với `x > 0=>` Áp dụng BĐT Cosi có: `\sqrt{x}+4/\sqrt{x} >= 2\sqrt{4}=4`
`<=>\sqrt{x}+1+4/\sqrt{x} >= 5`
Hay `S >= 5`
Dấu "`=`" xảy ra `<=>\sqrt{x}=4/\sqrt{x}<=>x=4` (t/m)
