`2)` Với `n\in N`* ta có:
`\qquad 1/{\sqrt{n}+\sqrt{n+1}}`
`={\sqrt{n+1}-\sqrt{n}}/{(\sqrt{n+1}+\sqrt{n})(\sqrt{n+1}-\sqrt{n})}`
`={\sqrt{n+1}-\sqrt{n}}/{n+1-n}=\sqrt{n+1}-\sqrt{n}`
`=>1/{\sqrt{n}+\sqrt{n+1}}=\sqrt{n+1}-\sqrt{n}` `(n\in N`*) $(1)$
Áp dụng $(1)$ ta có:
`A=1/{\sqrt{1}+\sqrt{2}}+1/{\sqrt{2}+\sqrt{3}}+1/{\sqrt{3}+\sqrt{4}}+...+1/{\sqrt{2021}+\sqrt{2022}}`
`=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{2022}-\sqrt{2021}`
`=\sqrt{2022}-\sqrt{1}=\sqrt{2022}-1`
Vậy `A=\sqrt{2022}-1`
$\\$
`3)` `(\sqrt{22}-\sqrt{2})\sqrt{6+\sqrt{11}}`
`=(\sqrt{11}-1).\sqrt{2}.\sqrt{6+\sqrt{11}}`
`=(\sqrt{11}-1).\sqrt{12+2\sqrt{11}}`
`=(\sqrt{11}-1).\sqrt{11+2.\sqrt{11}.1+1^2}`
`=(\sqrt{11}-1).\sqrt{(\sqrt{11}+1)^2}`
`=(\sqrt{11}-1).|\sqrt{11}+1|`
`=(\sqrt{11}-1)(\sqrt{11}+1)=11-1^2=10`
Vậy: `(\sqrt{22}-\sqrt{2})\sqrt{6+\sqrt{11}}=10`(đpcm)
