Đáp án:
`A=1/\sqrt{k}-1/\sqrt{k+1}` với `k> 0`
Giải thích các bước giải:
Ta có:
` A=1/{(k+1)\sqrt{k}+k\sqrt{k+1}}` `(k> 0)`
`A=1/{\sqrt{k}.\sqrt{k+1}.(\sqrt{k+1}+\sqrt{k})}`
`A={\sqrt{k+1}-\sqrt{k}}/{\sqrt{k}.\sqrt{k+1}.(\sqrt{k+1}+\sqrt{k}).(\sqrt{k+1}-\sqrt{k})}`
`A={\sqrt{k+1}-\sqrt{k}}/{\sqrt{k}.\sqrt{k+1}.[(\sqrt{k+1})^2-(\sqrt{k})^2]}`
`A={\sqrt{k+1}-\sqrt{k}}/{\sqrt{k}.\sqrt{k+1}.(k+1-k)}`
`A={\sqrt{k+1}-\sqrt{k}}/{\sqrt{k}.\sqrt{k+1}}`
`A=\sqrt{k+1}/{\sqrt{k}.\sqrt{k+1}}-\sqrt{k}/{\sqrt{k}.\sqrt{k+1}}`
`A=1/\sqrt{k}-1/\sqrt{k+1}`
Vậy: `A=1/\sqrt{k}-1/\sqrt{k+1}` với `k> 0`
