Đáp án:
`\sqrt{3}+\sqrt{2}`
Giải thích các bước giải:
`\qquad (1/{\sqrt{7-\sqrt{24}}+1}-1/{\sqrt{7+\sqrt{24}}-1}+1):(\sqrt{3}-\sqrt{2})`
`=(1/{\sqrt{6-2\sqrt{6}.1+1^2}+1}-1/{\sqrt{6+2\sqrt{6}.1+1^2}-1}):(\sqrt{3}-\sqrt{2})`
`=(1/{\sqrt{(\sqrt{6}-1)^2}+1}-1/{\sqrt{(\sqrt{6}+1)^2}-1}+1):(\sqrt{3}-\sqrt{2})`
`=(1/{\sqrt{6}-1+1}-1/{\sqrt{6}+1-1}+1):(\sqrt{3}-\sqrt{2})`
`=(1/\sqrt{6}-1/\sqrt{6}+1):(\sqrt{3}-\sqrt{2})`
`=1/{\sqrt{3}-\sqrt{2}}`
`={\sqrt{3}+\sqrt{2}}/{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}`
`={\sqrt{3}+\sqrt{2}}/{3-2}=\sqrt{3}+\sqrt{2}`
Vậy: `(1/{\sqrt{7-\sqrt{24}}+1}-1/{\sqrt{7+\sqrt{24}}-1}-1):(\sqrt{3}-\sqrt{2})=\sqrt{3}+\sqrt{2}`
