【数学I】例題1.2.4：2重根号（One More）★★

% 例題I1.2.4：2重根号 （One More）★★
次の2重根号を簡単な形にせよ. (1)$\sqrt{9+2\sqrt{14}}$(2)$\sqrt{7-2\sqrt{10}}$(3)$\sqrt{6+4\sqrt{2}}$(4)$\sqrt{8-\sqrt{48}}$(5)$\sqrt{4+\sqrt{15}}$

% 解答（例題I1.2.4）
(1)$\sqrt{9+2\sqrt{14}}=\sqrt{(7+2)+2\sqrt{7 \times 2}}=\sqrt{7}+\sqrt{2}$(2)$\sqrt{7-2\sqrt{10}}=\sqrt{(5+2)-2\sqrt{5 \times 2}}=\sqrt{5}-\sqrt{2}$(3)$\sqrt{6+4\sqrt{2}}=\sqrt{6+2\sqrt{8}}=\sqrt{(4+2)+2\sqrt{4 \times 2}}=2+\sqrt{2}$(4)$\sqrt{8-\sqrt{48}}=\sqrt{8-2\sqrt{12}}=\sqrt{(6+2)-2\sqrt{6 \times 2}}=\sqrt{6}-\sqrt{2}$(5)$\begin{aligned} \sqrt{4+\sqrt{15}}&=\sqrt{\dfrac{8+2\sqrt{15}}{2}}=\dfrac{\sqrt{(5+3)+2\sqrt{5\times 3}}}{\sqrt{2}}=\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}\\ &=\dfrac{\sqrt{2}(\sqrt{5}+\sqrt{3})}{2}=\dfrac{\sqrt{10}+\sqrt{6}}{2} \end{aligned}$

% 問題I1.2.4
次の2重根号を簡単な形にせよ. (1)$\sqrt{7-2\sqrt{12}}$(2)$\sqrt{12+6\sqrt{3}}$(3)$\sqrt{10-\sqrt{84}}$(4)$\sqrt{8+3\sqrt{7}}$

% 解答I1.2.4
(1)$\sqrt{7-2\sqrt{12}}=\sqrt{(4+3)-2\sqrt{4 \times 3}}=\sqrt{4}-\sqrt{3}=2-\sqrt{3}$(2)$\sqrt{12+6\sqrt{3}}=\sqrt{(9+3)+2\sqrt{9 \times 3}}=3+\sqrt{3}$(3)$\sqrt{10-\sqrt{84}}=\sqrt{10-2\sqrt{21}}=\sqrt{(7+3)-2\sqrt{7 \times 3}}=\sqrt{7}-\sqrt{3}$(4)$\begin{aligned} \sqrt{8+3\sqrt{7}}&=\sqrt{\dfrac{16+2\sqrt{63}}{2}}=\dfrac{\sqrt{(9+7)+2\sqrt{9 \times 7}}}{\sqrt{2}}\\ &=\dfrac{\sqrt{9}+\sqrt{7}}{\sqrt{2}}=\dfrac{3\sqrt{2}+\sqrt{14}}{2} \end{aligned}$