与えられた数学の問題は、根号を含む式の計算問題です。問題は全部で12問あります。

1. 問題の内容

与えられた数学の問題は、根号を含む式の計算問題です。問題は全部で12問あります。

2. 解き方の手順

(3) $(2\sqrt{2}-1)(2\sqrt{2}-3)$

まず、展開します。

$(2\sqrt{2}-1)(2\sqrt{2}-3) = (2\sqrt{2})(2\sqrt{2}) + (2\sqrt{2})(-3) + (-1)(2\sqrt{2}) + (-1)(-3)$

$= 4(2) - 6\sqrt{2} - 2\sqrt{2} + 3$

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

$= 11 - 8\sqrt{2}$

(4) $(\sqrt{5}+\sqrt{2})^2$

$(\sqrt{5}+\sqrt{2})^2 = (\sqrt{5})^2 + 2(\sqrt{5})(\sqrt{2}) + (\sqrt{2})^2$

$= 5 + 2\sqrt{10} + 2$

$= 7 + 2\sqrt{10}$

(5) $(\sqrt{2}-1)^2$

$(\sqrt{2}-1)^2 = (\sqrt{2})^2 - 2(\sqrt{2})(1) + (1)^2$

$= 2 - 2\sqrt{2} + 1$

$= 3 - 2\sqrt{2}$

(6) $(1-\sqrt{7})^2$

$(1-\sqrt{7})^2 = (1)^2 - 2(1)(\sqrt{7}) + (\sqrt{7})^2$

$= 1 - 2\sqrt{7} + 7$

$= 8 - 2\sqrt{7}$

(7) $(\sqrt{6}-\sqrt{2})^2$

$(\sqrt{6}-\sqrt{2})^2 = (\sqrt{6})^2 - 2(\sqrt{6})(\sqrt{2}) + (\sqrt{2})^2$

$= 6 - 2\sqrt{12} + 2$

$= 8 - 2\sqrt{4 \times 3}$

$= 8 - 2(2\sqrt{3})$

$= 8 - 4\sqrt{3}$

(8) $(\sqrt{7}+2)(\sqrt{7}-2)$

$(\sqrt{7}+2)(\sqrt{7}-2) = (\sqrt{7})^2 - (2)^2$

$= 7 - 4$

$= 3$

(9) $(3+\sqrt{5})(3-\sqrt{5})$

$(3+\sqrt{5})(3-\sqrt{5}) = (3)^2 - (\sqrt{5})^2$

$= 9 - 5$

$= 4$

(10) $(\sqrt{2}+\sqrt{6})(\sqrt{2}-\sqrt{6})$

$(\sqrt{2}+\sqrt{6})(\sqrt{2}-\sqrt{6}) = (\sqrt{2})^2 - (\sqrt{6})^2$

$= 2 - 6$

$= -4$

(11) $(2+\sqrt{2})(\sqrt{2}-1)$

$(2+\sqrt{2})(\sqrt{2}-1) = 2\sqrt{2} - 2 + 2 - \sqrt{2}$

$= \sqrt{2}$

(12) $(\sqrt{12}-\sqrt{8})(\sqrt{3}+\sqrt{2})$

$(\sqrt{12}-\sqrt{8})(\sqrt{3}+\sqrt{2}) = (2\sqrt{3}-2\sqrt{2})(\sqrt{3}+\sqrt{2})$

$= 2(\sqrt{3})^2 + 2\sqrt{6} - 2\sqrt{6} - 2(\sqrt{2})^2$

$= 2(3) - 2(2)$

$= 6 - 4$

$= 2$

3. 最終的な答え

(3) $11 - 8\sqrt{2}$

(4) $7 + 2\sqrt{10}$

(5) $3 - 2\sqrt{2}$

(6) $8 - 2\sqrt{7}$

(7) $8 - 4\sqrt{3}$

(8) $3$

(9) $4$

(10) $-4$

(11) $\sqrt{2}$

(12) $2$