# Quod Erat Demonstrandum

## 2012/07/25

### 根中根

Filed under: Additional / Applied Mathematics,HKCEE,NSS — johnmayhk @ 11:40 上午

$\sqrt{3+\sqrt{8}}=1+\sqrt{2}$

$\sqrt{3+\sqrt{7}}$

$\sqrt{x+\sqrt{y}}$

$\sqrt{x+\sqrt{y}}=\sqrt{a}+\sqrt{b}$ ………. (*)

$x+\sqrt{y}=(\sqrt{a}+\sqrt{b})^2=a+b+\sqrt{4ab}$

$a+b=x$
$4ab=y$

$4a^2-4ax+y=0$

$a=\frac{4x\pm \sqrt{16x^2-16y}}{8}=\frac{x\pm \sqrt{x^2-y}}{2}$ ………. (1)

$b=\frac{4x\mp \sqrt{16x^2-16y}}{8}=\frac{x\mp \sqrt{x^2-y}}{2}$ ………. (2)

$x^2-y=z^2$，其中 $z$ 是有理數。

$y=(x-z)(x+z)$

$\sqrt{2+\sqrt{3}}=\sqrt{1.5}+\sqrt{0.5}$
$\sqrt{3+\sqrt{8}}=\sqrt{2}+\sqrt{1}$
$\sqrt{4+\sqrt{15}}=\sqrt{2.5}+\sqrt{1.5}$
$\sqrt{5+\sqrt{24}}=\sqrt{3}+\sqrt{2}$
$\sqrt{3+\sqrt{5}}=\sqrt{2.5}+\sqrt{0.5}$
$\sqrt{4+\sqrt{12}}=\sqrt{3}+\sqrt{1}$
$\sqrt{5+\sqrt{21}}=\sqrt{3.5}+\sqrt{1.5}$
$\sqrt{6+\sqrt{32}}=\sqrt{4}+\sqrt{2}$
$\sqrt{5+\sqrt{16}}=\sqrt{4}+\sqrt{1}$（嗯… …）
$\sqrt{6+\sqrt{27}}=\sqrt{4.5}+\sqrt{1.5}$
$\sqrt{7+\sqrt{40}}=\sqrt{5}+\sqrt{2}$
$\sqrt{8+\sqrt{55}}=\sqrt{5.5}+\sqrt{2.5}$

$\sqrt{x+\sqrt{y}}=\sqrt{\frac{x+ \sqrt{x^2-y}}{2}}+\sqrt{\frac{x- \sqrt{x^2-y}}{2}}$

$\sqrt{3+\sqrt{7}}=\sqrt{\frac{3+ \sqrt{2}}{2}}+\sqrt{\frac{3- \sqrt{2}}{2}}$