Lời giải.
Ta có:
`+)3-2\sqrt{2}=2-2.\sqrt{2}.1+1=(\sqrt{2}-1)^2`
`=>\sqrt{3-2\sqrt{2}}=\sqrt{(\sqrt{2}-1)^2}=\sqrt{2}-1.`
Tương tự: `\sqrt{3+2\sqrt{2}}=\sqrt{(\sqrt{2}+1)^2}=\sqrt{2}+1.`
`+)17-12\sqrt{2}=8-2.3.2\sqrt{2}+9=8-2.3.\sqrt{2^2}.\sqrt{2}+9`
`=8-2.3.\sqrt{2^2 . 2}.\sqrt{2}+9=(\sqrt{8})^2- 2.3. \sqrt{8} + 3^2 = (\sqrt{8}-3)^2`
`=>\sqrt{17-12\sqrt{2}}=\sqrt{ (\sqrt{8}-3)^2}=3-\sqrt{8}=3-2\sqrt{2}.`
Tương tự: `17+12\sqrt{2}=3+\sqrt{8}=3+2\sqrt{2}.`
Trở lại biểu thức:
`\sqrt{3-2\sqrt{2}}/{\sqrt{17-12\sqrt{2}}}-\sqrt{3+2\sqrt{2}}/{\sqrt{17+12\sqrt{2}}}`
`=\sqrt{3-2\sqrt{2}}/{3-2\sqrt{2}}-\sqrt{3+2\sqrt{2}}/{3+2\sqrt{2}}`
`=\sqrt{3-2\sqrt{2}}/{\sqrt{3-2\sqrt{2}}.\sqrt{3-2\sqrt{2}}}-\sqrt{3+2\sqrt{2}}/{\sqrt{3+2\sqrt{2}}.\sqrt{3+2\sqrt{2}}}`
`=1/{\sqrt{3-2\sqrt{2}}}- 1/{\sqrt{3+2\sqrt{2}}}`
`=1/{\sqrt{2}-1}- 1/{\sqrt{2}+1}`
`={\sqrt{2}+1-\sqrt{2}+1}/{(\sqrt{2}-1).(\sqrt{2}+1)}`
`=2/{2-1}=2.`
