与えられた数の平方根を、できる限り簡単な形で表す問題です。

1. 問題の内容

与えられた数の平方根を、できる限り簡単な形で表す問題です。

2. 解き方の手順

平方根の中身を素因数分解し、$a^2$ の形を見つけて、根号の外に出します。

$\sqrt{52} = \sqrt{2^2 \cdot 13} = 2\sqrt{13}$

$\sqrt{54} = \sqrt{3^2 \cdot 6} = 3\sqrt{6}$

$\sqrt{56} = \sqrt{2^3 \cdot 7} = \sqrt{2^2 \cdot 2 \cdot 7} = 2\sqrt{14}$

$\sqrt{60} = \sqrt{2^2 \cdot 3 \cdot 5} = 2\sqrt{15}$

$\sqrt{63} = \sqrt{3^2 \cdot 7} = 3\sqrt{7}$

$\sqrt{68} = \sqrt{2^2 \cdot 17} = 2\sqrt{17}$

$\sqrt{72} = \sqrt{2^3 \cdot 3^2} = \sqrt{2^2 \cdot 3^2 \cdot 2} = 2 \cdot 3 \sqrt{2} = 6\sqrt{2}$

$\sqrt{76} = \sqrt{2^2 \cdot 19} = 2\sqrt{19}$

$\sqrt{80} = \sqrt{2^4 \cdot 5} = \sqrt{(2^2)^2 \cdot 5} = 2^2 \sqrt{5} = 4\sqrt{5}$

$\sqrt{84} = \sqrt{2^2 \cdot 3 \cdot 7} = 2\sqrt{21}$

$\sqrt{88} = \sqrt{2^3 \cdot 11} = \sqrt{2^2 \cdot 2 \cdot 11} = 2\sqrt{22}$

$\sqrt{90} = \sqrt{2 \cdot 3^2 \cdot 5} = 3\sqrt{10}$

$\sqrt{92} = \sqrt{2^2 \cdot 23} = 2\sqrt{23}$

$\sqrt{96} = \sqrt{2^5 \cdot 3} = \sqrt{2^4 \cdot 2 \cdot 3} = \sqrt{(2^2)^2 \cdot 6} = 4\sqrt{6}$

$\sqrt{98} = \sqrt{2 \cdot 7^2} = 7\sqrt{2}$

$\sqrt{99} = \sqrt{3^2 \cdot 11} = 3\sqrt{11}$

$\sqrt{100} = 10$

3. 最終的な答え

$\sqrt{52} = 2\sqrt{13}$

$\sqrt{54} = 3\sqrt{6}$

$\sqrt{56} = 2\sqrt{14}$

$\sqrt{60} = 2\sqrt{15}$

$\sqrt{63} = 3\sqrt{7}$

$\sqrt{68} = 2\sqrt{17}$

$\sqrt{72} = 6\sqrt{2}$

$\sqrt{76} = 2\sqrt{19}$

$\sqrt{80} = 4\sqrt{5}$

$\sqrt{84} = 2\sqrt{21}$

$\sqrt{88} = 2\sqrt{22}$

$\sqrt{90} = 3\sqrt{10}$

$\sqrt{92} = 2\sqrt{23}$

$\sqrt{96} = 4\sqrt{6}$

$\sqrt{98} = 7\sqrt{2}$

$\sqrt{99} = 3\sqrt{11}$

$\sqrt{100} = 10$