数学演習/中学校3年生/平方根/解答のソースを表示

[[数学演習 中学校3年生]]

[[中学校数学 3年生-数量/平方根]]

問題は[[数学演習 中学校3年生 平方根|こちら]]にあります。

== 平方根(1) ==
#<math>\sqrt{36-11} = \sqrt{25} = \sqrt{5^2} = 5</math>
#<math>\sqrt{121} = \sqrt{11^2} = 11</math>
#<math>\sqrt{\frac{169}{49}} = \sqrt{\frac{13^2}{7^2}} = \frac{\sqrt{13^2}}{\sqrt{7^2}} = \frac{13}{7}</math>
#<math>\sqrt{x^4} = \sqrt{(x^2)^2} = x^2</math>
#<math>\sqrt{\frac{x^2}{y^2}} = \frac{\sqrt{x^2}}{\sqrt{y^2}} = \frac{x}{y}</math>
#<math>\sqrt{\frac{4x^4}{225y^2z^8}} = \frac{\sqrt{2^2 \times (x^2)^2}}{\sqrt{15^2 \times y^2 \times (z^4)^2}} = \frac{2x^2}{15yz^4}</math>
#<math>\sqrt{36x^2+96xy+64y^2} = \sqrt{(6x)^2+2(6x \times 8y)+(8y)^2} =\sqrt{(6x+8y)^2}=6x+8y</math>

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

== 平方根(2) ==
#<math>\sqrt{12} = \sqrt{3 \times 4} = \sqrt{3 \times 2^2} = 2 \sqrt{3}</math>
#<math>\sqrt{75} = \sqrt{3 \times 25} = \sqrt{3 \times 5^2} = 5 \sqrt{3}</math>
#<math>\sqrt{396} = \sqrt{36 \times 11} = \sqrt{6^2 \times 11} = 6 \sqrt{11}</math>
#<math>\sqrt{(x+y)^3} = \sqrt{(x+y) \times (x+y)^2} = (x+y) \sqrt{x+y}</math>
#<math>6\sqrt{396}+\sqrt{1331} = 6\sqrt{36 \times 11}+\sqrt{121 \times 11} = (36+11)\sqrt{11} = 47\sqrt{11}</math>

== 平方根の計算(1) ==
#<math>\sqrt{2}+2\sqrt{2} = (1+2)\sqrt{2} = 3 \sqrt{2}</math>
#<math>\sqrt{5}-5\sqrt{5} = (1-5)\sqrt{5} = -4 \sqrt{5}</math>
#<math>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}</math>
#<math>\sqrt{3} \times 3 \sqrt{7} = 3 \sqrt{3 \times 7} = 3 \sqrt{21}</math>
#<math>\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}</math>
#<math>(\sqrt{2}+2)(1+\sqrt{3}) = \sqrt{2}+\sqrt{2 \times 3}+2+2\sqrt{3} = 2+2\sqrt{3}+\sqrt{6}+\sqrt{2}</math>
#<math>(\sqrt{5}-2)^2 = (\sqrt{5})^2-2\times \sqrt{5}\times 2 +2^2 = 9-4\sqrt{5}</math>
#<math>(\sqrt{7}+\sqrt{3})(\sqrt{7}-\sqrt{3}) = (\sqrt{7})^2-(\sqrt{3})^2 = 7-3 = 4</math>
#<math>(6+\sqrt{55})(6-\sqrt{55}) = 6^2-(\sqrt{55})^2 = 36-55 = -19</math>

== 平方根の計算(2) ==
#<math>\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}</math>
#<math>\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} </math>
#<math>\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}</math>
#<math>\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}</math>

[[カテゴリ:中学校数学演習|3年へいほうこんこたえ]]