ĐKXĐ: `x\geq0,x\ne1`
`a,` `x=7-4\sqrt{3}=7-2\sqrt{12}=4-2\sqrt{12}+3=(\sqrt{4}-\sqrt{3})^2=(2-\sqrt{3})^2` (Thỏa mãn)
Thay `x=(2-\sqrt{3})^2` vào `Q` có:
`Q={\sqrt{(2-\sqrt{3})^2}+1}/{\sqrt{(2-\sqrt{3})^2}+2}`
`={|2-\sqrt{3}|+1}/{|2-\sqrt{3}|+2}`
`={2-\sqrt{3}+1}/{2-\sqrt{3}+2}`
`={3-\sqrt{3}}/{4-\sqrt{3}}`
`={(3-\sqrt{3})(4+\sqrt{3})}/{(4+\sqrt{3})(4-\sqrt{3})}`
`={12+3\sqrt{3}-4\sqrt{3}-3}/{16-3}`
`={9-\sqrt{3}}/{13}`
Vậy với `x=7-4\sqrt{3}` thì `Q={9-\sqrt{3}}/{13}`
`b,`
`P={\sqrt{x}}/{\sqrt{x}+1}-{1}/{1-\sqrt{x}}-{2\sqrt{x}}/{x-1}`
`={\sqrt{x}}/{\sqrt{x}+1}+{1}/{\sqrt{x}-1}-{2\sqrt{x}}/{(\sqrt{x}+1)(\sqrt{x}-1)}`
`={\sqrt{x}(\sqrt{x}-1)}/{(\sqrt{x}+1)(\sqrt{x}-1)}+{\sqrt{x}+1}/{(\sqrt{x}+1)(\sqrt{x}-1)}-{2\sqrt{x}}/{(\sqrt{x}+1)(\sqrt{x}-1)}`
`={\sqrt{x}(\sqrt{x}-1)+\sqrt{x}+1-2\sqrt{x}}/{(\sqrt{x}+1)(\sqrt{x}-1)}`
`={x-\sqrt{x}+\sqrt{x}+1-2\sqrt{x}}/{(\sqrt{x}+1)(\sqrt{x}-1)}`
`={x-2\sqrt{x}+1}/{(\sqrt{x}+1)(\sqrt{x}-1)}`
`={(\sqrt{x}-1)^2}/{(\sqrt{x}+1)(\sqrt{x}-1)}`
`={\sqrt{x}-1}/{\sqrt{x}+1}`
`M=P.Q`
`⇒M={\sqrt{x}-1}/{\sqrt{x}+1}.{\sqrt{x}+1}/{\sqrt{x}+2}={\sqrt{x}-1}/{\sqrt{x}+2}`
Vậy `x\geq0,x\ne1` thì `M={\sqrt{x}-1}/{\sqrt{x}+2}`
`c,`
`M<{-1}/{3}`
`⇒{\sqrt{x}-1}/{\sqrt{x}+2}<{-1}/{3}`
`⇔{\sqrt{x}-1}/{\sqrt{x}+2}+{1}/{3}<0`
`⇔{3(\sqrt{x}-1)}/{3(\sqrt{x}+2)}+{\sqrt{x}+2}/{3(\sqrt{x}+2)}<0`
`⇔{3(\sqrt{x}-1)+\sqrt{x}+2}/{3(\sqrt{x}+2)}<0`
`⇔{3\sqrt{x}-3+\sqrt{x}+2}/{3(\sqrt{x}+2)}<0`
`⇔{4\sqrt{x}-1}/{3(\sqrt{x}+2)}<0`
Có `\sqrt{x}\geq0` với mọi `x\geq0`
`⇒\sqrt{x}+2\geq2>0` với mọi `x\geq0`
`⇒3(\sqrt{x}+2)\geq6>0` với mọi `x\geq0`
`M<{-1}/{3}`
`⇔4\sqrt{x}-1<0`
`⇔4\sqrt{x}<1`
`⇔\sqrt{x}<1/4`
`⇔x<1/16`
Kết hợp ĐKXĐ: `x\geq0`
`⇒0\leqx<1/16`
Vậy với `0\leqx<1/16` thì `M<{-1}/{3}`
`d,`
Có `\sqrt{x}\geq0` với mọi `x\geq0`
`⇒\sqrt{x}-1\geq-1` với mọi `x\geq0`
`\sqrt{x}+2\geq2` với mọi `x\geq0`
`⇒{1}/{\sqrt{x}+2}\leq{1}/{2}` với mọi `x\geq0`
`⇒{\sqrt{x}-1}/{\sqrt{x}+2}\geq{-1}/{2}` với mọi `x\geq0`
$⇒MinM=$`{-1}/{2}`.
Dấu "=" xảy ra `⇔\sqrt{x}=0⇔x=0` (Thỏa mãn)
Vậy với `x=0` thì $MinM=$`{-1}/{2}`
