`ĐKXĐ:x>0`
`a,A={1}/{3-\sqrt{5}}-{1}/{1+\sqrt{5}}`
`={3+\sqrt{5}}/{(3+\sqrt{5})(3-\sqrt{5})}-{1-\sqrt{5}}/{(1+\sqrt{5})(1-\sqrt{5})}`
`={3+\sqrt{5}}/{9-5}-{1-\sqrt{5}}/{1-5}`
`={3+\sqrt{5}}/{4}-{1-\sqrt{5}}/{-4}`
`={3+\sqrt{5}}/{4}+{1-\sqrt{5}}/{4}`
`={3+\sqrt{5}+1-\sqrt{5}}/{4}`
`={4}/{4}`
`=1`
`B={x+\sqrt{x}}/{\sqrt{x}}+{x-4}/{\sqrt{x}+2}`
`={\sqrt{x}(\sqrt{x}+1)}/{\sqrt{x}}+{(\sqrt{x}+2)(\sqrt{x}-2)}/{\sqrt{x}+2}`
`=\sqrt{x}+1+\sqrt{x}-2`
`=2\sqrt{x}-1`
Vậy với `x>0` thì `A=1` và `B=2\sqrt{x}-1`
`b,A+B=2`
`⇔1+2\sqrt{x}-1=2`
`⇔2\sqrt{x}=2`
`⇔\sqrt{x}=1`
`⇔x=1(TM)`
Vậy với `x=1` thì `A+B=2`