`ĐKXĐ:x\ge0,x\ne4`
`1,x=7+4\sqrt{3}=7+2\sqrt{12}=4+2\sqrt{12}+3=(\sqrt{4}+\sqrt{3})^2=(2+\sqrt{3})^2(TM)`
Thay `x=(2+\sqrt{3})^2` vào `A` có:
`A={\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}}/{\sqrt{3}}`
`={\sqrt{3}(\sqrt{3}+1)}/{\sqrt{3}}`
`=\sqrt{3}+1`
Vậy với `x=7+4\sqrt{3}` thì `A=\sqrt{3}+1`
`2,x-\sqrt{x}-2=x+\sqrt{x}-2\sqrt{x}-2=\sqrt{x}(\sqrt{x}+1)-2(\sqrt{x}+1)=(\sqrt{x}+1)(\sqrt{x}-2)`
`B={\sqrt{x}}/{\sqrt{x}+1}+{1-\sqrt{x}}/{\sqrt{x}-2}-{\sqrt{x}+4}/{x-\sqrt{x}-2}`
`={\sqrt{x}}/{\sqrt{x}+1}+{1-\sqrt{x}}/{\sqrt{x}-2}-{\sqrt{x}+4}/{(\sqrt{x}+1)(\sqrt{x}-2)}`
`={\sqrt{x}(\sqrt{x}-2)}/{(\sqrt{x}+1)(\sqrt{x}-2)}+{(1-\sqrt{x})(\sqrt{x}+1)}/{(\sqrt{x}+1)(\sqrt{x}-2)}-{\sqrt{x}+4}/{(\sqrt{x}+1)(\sqrt{x}-2)}`
`={\sqrt{x}(\sqrt{x}-2)+(1-\sqrt{x})(\sqrt{x}+1)-(\sqrt{x}+4)}/{(\sqrt{x}+1)(\sqrt{x}-2)}`
`={x-2\sqrt{x}+1-x-\sqrt{x}-4}/{(\sqrt{x}+1)(\sqrt{x}-2)}`
`={-3\sqrt{x}-3}/{(\sqrt{x}+1)(\sqrt{x}-2)}`
`={-3(\sqrt{x}+1)}/{(\sqrt{x}+1)(\sqrt{x}-2)}`
`={-3}/{\sqrt{x}-2}`
`={3}/{2-\sqrt{x}}`
Vậy với `x\ge0,x\ne4` thì `B={3}/{2-\sqrt{x}}`
