## 2019年10月16日水曜日

### 数学 - 代数学 - 実数 - 無理数、平方根、簡約

1. $\begin{array}{l}2\sqrt{75}-\sqrt{108}-\frac{\sqrt{27}}{2}+\frac{5}{\sqrt{12}}\\ =10\sqrt{3}-6\sqrt{3}-\frac{3\sqrt{3}}{2}+\frac{10\sqrt{3}}{12}\\ =\frac{5\sqrt{3}}{2}+\frac{5\sqrt{3}}{6}\\ =\frac{20\sqrt{3}}{6}\\ =\frac{10\sqrt{3}}{3}\end{array}$

2. $\begin{array}{l}{\left(3+2\sqrt{2}\right)}^{2}-{\left(3-2\sqrt{2}\right)}^{2}\\ =\left(\left(3+2\sqrt{2}\right)+\left(3-2\sqrt{2}\right)\right)\left(3+2\sqrt{2}\right)-\left(3-2\sqrt{2}\right)\right)\\ =6·4\sqrt{2}\\ =24\sqrt{2}\end{array}$

3. $\begin{array}{l}3{\left(\frac{2-\sqrt{10}}{3}\right)}^{2}-4\left(\frac{2-\sqrt{10}}{3}\right)-2\\ =\left(\left(2-\sqrt{10}\right)-4\right)\left(\frac{2-\sqrt{10}}{3}\right)-2\\ =\frac{-\left(2+\sqrt{10}\right)\left(2-\sqrt{10}\right)}{3}-2\\ =\frac{-4+10-6}{3}\\ =0\end{array}$

4. $\begin{array}{l}\frac{\sqrt{2}}{\sqrt{3}+\sqrt{2}}-\frac{\sqrt{3}}{\sqrt{3}-\sqrt{2}}\\ =\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)\\ =-2-3\\ =-5\end{array}$

5. $\begin{array}{l}\frac{3\sqrt{7}-7\sqrt{3}}{3\sqrt{7}+7\sqrt{3}}\\ =\frac{{3}^{2}·7-2·7·3\sqrt{21}+{7}^{2}·3}{{3}^{2}·7-{7}^{2}·3}\\ =\frac{3-2\sqrt{21}+7}{3-7}\\ =\frac{10-2\sqrt{21}}{-4}\\ =\frac{\sqrt{21}-5}{2}\end{array}$

6. $\begin{array}{l}\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}+\frac{3\sqrt{2}}{3+\sqrt{6}}-\frac{4\sqrt{3}}{\sqrt{2}+\sqrt{6}}\\ =-\sqrt{6}\left(\sqrt{2}-\sqrt{3}\right)+\frac{3\sqrt{2}\left(3-\sqrt{6}\right)}{3}-\frac{4\sqrt{3}\left(\sqrt{2}-\sqrt{6}\right)}{-4}\\ =-2\sqrt{3}+3\sqrt{2}+3\sqrt{2}-2\sqrt{3}+\sqrt{6}-3\sqrt{2}\\ =3\sqrt{2}-4\sqrt{3}+\sqrt{6}\end{array}$

7. $\begin{array}{l}\sqrt{\frac{2}{6+\sqrt{35}}}\\ =\sqrt{2\left(6-\sqrt{35}\right)}\\ =\sqrt{12-2\sqrt{35}}\\ =\sqrt{{\left(\sqrt{7}-\sqrt{5}\right)}^{2}}\\ =\sqrt{7}-\sqrt{5}\end{array}$

8. $\begin{array}{l}\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}-\sqrt{\frac{\sqrt{2}+1}{\sqrt{2}-1}}\\ =\sqrt{3-2\sqrt{2}}-\sqrt{3+2\sqrt{2}}\\ =\left(\sqrt{2}-1\right)-\left(\sqrt{2}+1\right)\\ =-2\end{array}$

9. $\begin{array}{l}\sqrt{7-3\sqrt{5}}+\sqrt{3+\sqrt{5}}\\ =\sqrt{\frac{14-2\sqrt{45}}{2}}+\sqrt{\frac{6+2\sqrt{5}}{2}}\\ =\frac{3-\sqrt{5}}{\sqrt{2}}+\frac{\sqrt{5}+1}{\sqrt{2}}\\ =\frac{4}{\sqrt{2}}\\ =2\sqrt{2}\end{array}$