计算:
\[
\dfrac{
1 + \sqrt{2-\sqrt2} + \sqrt{2-\sqrt3}
}{
\sqrt3 + \sqrt{2+\sqrt2} + \sqrt{2+\sqrt3}
}
\]
Qer: zhaozhe 20180807
解答
\[
\begin{aligned}
\text{原式}
&= \dfrac{
1 + \sqrt{2-\sqrt2} + \dfrac{\sqrt6-\sqrt2}2
}{
\sqrt3 + \sqrt{2+\sqrt2} + \dfrac{\sqrt6+\sqrt2}2
} \\
&= \dfrac{
2\sqrt{2-\sqrt2} + 2 + \sqrt6 - \sqrt2
}{
2\sqrt{2+\sqrt2} + 2\sqrt3 + \sqrt6 + \sqrt2
}
\end{aligned}
\]
不难验证,
\[
(2\sqrt3 + \sqrt6 + \sqrt2) (\sqrt2 - 1) = 2 + \sqrt6 - \sqrt2
\]
以及
\[
\dfrac{\sqrt{2-\sqrt2}}{\sqrt{2+\sqrt2}} =
\sqrt{\dfrac{2-\sqrt2}{2+\sqrt2}} = \sqrt2 - 1
\]
因此, 原式 \(=\sqrt2-1\).
