Chứng minh rằng: c) `23-8 \sqrt{7}=(4-\sqrt{7})^{2}` d) `\sqrt{17-12 \sqrt{2}}+2 \sqrt{2}=3`

`c)` Ta có:

`23 - 8\sqrt{7}`

`= 16 - 2 . 16 .\sqrt{7} + 7`

`= 4^2 - 2 . 16 .\sqrt{7} + (\sqrt{7})^2`

`= (4 - \sqrt{7})^2` 

Vậy `23 - 8\sqrt{7} = (4 - \sqrt{7})^2` $\rightarrow$ (Đpcm)

`d)` Ta có:

`\sqrt{17 - 12\sqrt{2}} + 2\sqrt{2}`

`= \sqrt{9 - 2 . 3 . 2\sqrt{2} + 8} + 2\sqrt{2}`

`= \sqrt{3^2 - 2 . 3 . 2\sqrt{2} + (2\sqrt{2})^2} + 2\sqrt{2}`

`= \sqrt{(3 - 2\sqrt{2})^2} + 2\sqrt{2}`

`= (3 - 2\sqrt{2}) + 2\sqrt{2}`

`= 3 - 2\sqrt{2} + 2\sqrt{2}`

`= 3`

Vậy `\sqrt{17 - 12\sqrt{2}} + 2\sqrt{2} = 3` $\rightarrow$ (Đpcm)