数学演習/中学校3年生/平方根/解答

問題はこちらにあります。

平方根(1)

$\sqrt{36 - 11} = \sqrt{25} = \sqrt{5^{2}} = 5$
$\sqrt{121} = \sqrt{1 1^{2}} = 11$
$\sqrt{\frac{169}{49}} = \sqrt{\frac{1 3^{2}}{7^{2}}} = \frac{\sqrt{1 3^{2}}}{\sqrt{7^{2}}} = \frac{13}{7}$
$\sqrt{x^{4}} = \sqrt{(x^{2})^{2}} = x^{2}$
$\sqrt{\frac{x^{2}}{y^{2}}} = \frac{\sqrt{x^{2}}}{\sqrt{y^{2}}} = \frac{x}{y}$
$\sqrt{\frac{4 x^{4}}{225 y^{2} z^{8}}} = \frac{\sqrt{2^{2} \times (x^{2})^{2}}}{\sqrt{1 5^{2} \times y^{2} \times (z^{4})^{2}}} = \frac{2 x^{2}}{15 y z^{4}}$
$\sqrt{36 x^{2} + 96 x y + 64 y^{2}} = \sqrt{(6 x)^{2} + 2 (6 x \times 8 y) + (8 y)^{2}} = \sqrt{(6 x + 8 y)^{2}} = 6 x + 8 y$

7.は乗法公式 $a^{2} + 2 a b + b^{2} = (a + b)^{2}$ に $a = 6 x, b = 8 y$ を代入した形になっている。（実際に計算してみよう）

$\sqrt{12} = \sqrt{3 \times 4} = \sqrt{3 \times 2^{2}} = 2 \sqrt{3}$
$\sqrt{75} = \sqrt{3 \times 25} = \sqrt{3 \times 5^{2}} = 5 \sqrt{3}$
$\sqrt{396} = \sqrt{36 \times 11} = \sqrt{6^{2} \times 11} = 6 \sqrt{11}$
$\sqrt{(x + y)^{3}} = \sqrt{(x + y) \times (x + y)^{2}} = (x + y) \sqrt{x + y}$
$6 \sqrt{396} + \sqrt{1331} = 6 \sqrt{36 \times 11} + \sqrt{121 \times 11} = (36 + 11) \sqrt{11} = 47 \sqrt{11}$

$\sqrt{2} + 2 \sqrt{2} = (1 + 2) \sqrt{2} = 3 \sqrt{2}$
$\sqrt{5} - 5 \sqrt{5} = (1 - 5) \sqrt{5} = - 4 \sqrt{5}$
$7 + 2 \sqrt{2} + 3 \sqrt{3} + 6 \sqrt{12} = 7 + 2 \sqrt{2} + 3 \sqrt{3} + 6 \sqrt{3 \times 2^{2}} = 7 + 2 \sqrt{2} + 3 \sqrt{3} + 12 \sqrt{3} = 7 + 2 \sqrt{2} + 15 \sqrt{3}$
$\sqrt{3} \times 3 \sqrt{7} = 3 \sqrt{3 \times 7} = 3 \sqrt{21}$
$\sqrt{6} \times 2 \sqrt{3} = \sqrt{2 \times 3} \times 2 \sqrt{3} = 2 \sqrt{2 \times 3 \times 3} = 2 \sqrt{2 \times 3^{2}} = 6 \sqrt{2}$
$(\sqrt{2} + 2) (1 + \sqrt{3}) = \sqrt{2} + \sqrt{2 \times 3} + 2 + 2 \sqrt{3} = 2 + 2 \sqrt{3} + \sqrt{6} + \sqrt{2}$
$(\sqrt{5} - 2)^{2} = (\sqrt{5})^{2} - 2 \times \sqrt{5} \times 2 + 2^{2} = 9 - 4 \sqrt{5}$
$(\sqrt{7} + \sqrt{3}) (\sqrt{7} - \sqrt{3}) = (\sqrt{7})^{2} - (\sqrt{3})^{2} = 7 - 3 = 4$
$(6 + \sqrt{55}) (6 - \sqrt{55}) = 6^{2} - (\sqrt{55})^{2} = 36 - 55 = - 19$

$\frac{2}{\sqrt{3}} = \frac{2}{\sqrt{3}} \times 1 = \frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{2 \sqrt{3}}{\sqrt{3^{2}}} = \frac{2 \sqrt{3}}{3}$
$\frac{\sqrt{6}}{\sqrt{5}} = \frac{\sqrt{6}}{\sqrt{5}} \times 1 = \frac{\sqrt{6}}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{6 \times 5}}{\sqrt{5^{2}}} = \frac{\sqrt{30}}{5}$
$\frac{3 + \sqrt{2}}{\sqrt{3}} = \frac{3 + \sqrt{2}}{\sqrt{3}} \times 1 = \frac{3 + \sqrt{2}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3} (3 + \sqrt{2})}{\sqrt{3^{2}}} = \frac{3 \sqrt{3} + \sqrt{2 \times 3}}{3} = \frac{3 \sqrt{3} + \sqrt{6}}{3}$
$\frac{\sqrt{2} + 1}{\sqrt{2} - 1} = \frac{\sqrt{2} + 1}{\sqrt{2} - 1} \times 1 = \frac{\sqrt{2} + 1}{\sqrt{2} - 1} \times \frac{\sqrt{2} + 1}{\sqrt{2} + 1} = \frac{(\sqrt{2} + 1)^{2}}{(\sqrt{2} - 1) (\sqrt{2} + 1)} = \frac{(\sqrt{2})^{2} + 2 \times \sqrt{2} \times 1 + 1^{2}}{(\sqrt{2})^{2} - 1^{2}} = \frac{3 + 2 \sqrt{2}}{1} = 3 + 2 \sqrt{2}$