【数学I】例題1.2.2：平方根の計算（One More）★

% 例題I1.2.2：平方根の計算 （One More）★
次の式を計算せよ. (1)$4\sqrt{12}-\sqrt{18}-2\sqrt{75}$(2)$(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})$(3)$(\sqrt{5}+\sqrt{3})^2+(\sqrt{5}-\sqrt{3})^2$(4)$(\sqrt{2}+\sqrt{3}+\sqrt{5})(\sqrt{2}+\sqrt{3}-\sqrt{5})$

% 解答（例題I1.2.2）
(1)$\begin{aligned} 4\sqrt{12}-\sqrt{18}-2\sqrt{75} &=4\sqrt{2^2 \cdot 3}-\sqrt{3^2 \cdot 2}-2\sqrt{5^2 \cdot 3}\\ &=8\sqrt{3}-3\sqrt{2}-10\sqrt{3}=-3\sqrt{2}-2\sqrt{3} \end{aligned}$(2)$(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})=(\sqrt{3})^2-(\sqrt{2})^2=3-2=1$(3)$\begin{aligned} (\sqrt{5}+\sqrt{3})^2+(\sqrt{5}-\sqrt{3})^2 &=(\sqrt{5})^2+2\sqrt{5}\sqrt{3}+(\sqrt{3})^2\\ &+(\sqrt{5})^2-2\sqrt{5}\sqrt{3}+(\sqrt{3})^2\\ &=5+2\sqrt{15}+3+5-2\sqrt{15}+3=16 \end{aligned}$(4)$\begin{aligned} (\sqrt{2}+\sqrt{3}+\sqrt{5})(\sqrt{2}+\sqrt{3}-\sqrt{5})&=(\sqrt{2}+\sqrt{3})^2-(\sqrt{5})^2\\ &=2+2\sqrt{6}+3-5=2\sqrt{6} \end{aligned}$

% 問題I1.2.2
次の式を計算せよ. (1)$3\sqrt{20}-2\sqrt{45}+\sqrt{27}$(2)$(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})$(3)$(\sqrt{7}+\sqrt{2})^2-(\sqrt{7}-\sqrt{2})^2$(4)$(\sqrt{5}+\sqrt{7}+\sqrt{3})(\sqrt{5}+\sqrt{7}-\sqrt{3})$

% 解答I1.2.2
(1)$\begin{aligned} 3\sqrt{20}-2\sqrt{45}+\sqrt{27} &=3\sqrt{2^2 \cdot 5}-2\sqrt{3^2 \cdot 5}+\sqrt{3^3}\\ &=6\sqrt{5}-6\sqrt{5}+3\sqrt{3}=3\sqrt{3} \end{aligned}$(2)$(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})=(\sqrt{6})^2-(\sqrt{2})^2=6-2=4$(3)$\begin{aligned} (\sqrt{7}+\sqrt{2})^2-(\sqrt{7}-\sqrt{2})^2 &=(\sqrt{7})^2+2\sqrt{7}\sqrt{2}+(\sqrt{2})^2\\ &-\{(\sqrt{7})^2-2\sqrt{7}\sqrt{2}+(\sqrt{2})^2\}\\ &=7+2\sqrt{14}+2-(7-2\sqrt{14}+2)=4\sqrt{14} \end{aligned}$(4)$\begin{aligned} (\sqrt{5}+\sqrt{7}+\sqrt{3})(\sqrt{5}+\sqrt{7}-\sqrt{3})&=(\sqrt{5}+\sqrt{7})^2-(\sqrt{3})^2\\ &=5+2\sqrt{35}+7-3=9+2\sqrt{35} \end{aligned}$