【数学１】数と式--実数 - かぽなーた〜高校数学オンライン問題集〜

実数

実数

今回のテーマ

今回のテーマは，実数です．

有理数と無理数
絶対値
平方根
分数の有理化

を扱っています．

授業動画はこちらをご覧ください

www.youtube.com

演習問題

${\fbox{1} \ 次の分数を循環小数に，また循環小数を分数にせよ。\\ ( 1 )\ \dfrac{1}{3}\\ ( 2 )\ \dfrac{7}{110}\\ ( 3 )\ 0.\dot{2}\dot{5}\\ ( 4 )\ 1.\dot{2}0\dot{4}\\ }$

${\fbox{2} \ 次の計算をせよ。\\ ( 1 )\ \sqrt{18}\times\sqrt{40}\\ ( 2 )\ 5\sqrt{8}-\sqrt{98}+\sqrt{32}\\ ( 3 )\ (2\sqrt{2}-\sqrt{3})(\sqrt{2}+2\sqrt{3})\\ ( 4 )\ (1+\sqrt{2}-\sqrt{3})^2\\ }$

${\fbox{3} \ 次の式を簡単にせよ。\\ ( 1 )\ \dfrac{1}{2\sqrt{2}}\\ ( 2 )\ \dfrac{\sqrt{5}}{\sqrt{5}-\sqrt{2}}\\ ( 3 )\ \dfrac{2+\sqrt{3}}{2-\sqrt{3}}\\ ( 4 )\ \dfrac{1}{3-\sqrt{2}+\sqrt{7}}\\ }$

${\fbox{4} \ 次の二重根号をはずせ。\\ ( 1 )\ \sqrt{6+2\sqrt{5}}\\ ( 2 )\ \sqrt{7-2\sqrt{12}}\\ ( 3 )\ \sqrt{6+\sqrt{32}}\\ ( 4 )\ \sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\\ }$

CheatSheet

f:id:brian_tee:20190403222218p:plain — cheat sheet

演習解答

${\fbox{1}\\ ( 1 )\ \dfrac{1}{3} = 0.\dot{3}\\ ( 2 )\ \dfrac{7}{110} = 0.0\dot{6}\dot{3}\\ ( 3 )\ 0.\dot{2}\dot{5}\\ \begin{align} \ x &= 0.2525...\ と置くと，\\ \ 100x &= 25.2525...\ なので，\\ \ 99x &= 25\\ \ x &= \dfrac{25}{99}\\ ( 4 )\ 1.\dot{2}0\dot{4}\\ \ x &= 1.204204...\ と置くと，\\ \ 1000x &= 1204.204204...\ なので，\\ \ 999x &= 1203\\ \ x &= \dfrac{1203}{999}\\ \ &= \dfrac{401}{333}\\ \end{align} \\ }$

${\fbox{2} \ \\ ( 1 )\ \sqrt{18}\times\sqrt{40}\\ \ = 3\sqrt{2}\times2\sqrt{10}\\ \ = 6\sqrt{20}\\ \ = 12\sqrt{5}\\ ( 2 )\ 5\sqrt{8}-\sqrt{98}+\sqrt{32}\\ \ =10\sqrt{2}-7\sqrt{2}+4\sqrt{2}\\ \ =7\sqrt{2}\\ ( 3 )\ (2\sqrt{2}-\sqrt{3})(\sqrt{2}+2\sqrt{3})\\ \ = 4-6+4\sqrt{6}-\sqrt{6}\\ \ = 3\sqrt{6}-2\\ ( 4 )\ (1+\sqrt{2}-\sqrt{3})^2\\ \ = 1+2+3+2\sqrt{2}-2\sqrt{3}-2\sqrt{6}\\ \ = 6+2\sqrt{2}-2\sqrt{3}-2\sqrt{6}\\ }$

${\fbox{3} \\ ( 1 )\ \dfrac{1}{2\sqrt{2}}\\ \ = \dfrac{1}{2\sqrt{2}}\times\dfrac{\sqrt{2}}{\sqrt{2}}\\ \ =\dfrac{\sqrt{2}}{4}\\ ( 2 )\ \dfrac{\sqrt{5}}{\sqrt{5}-\sqrt{2}}\\ \ =\dfrac{\sqrt{5}}{\sqrt{5}-\sqrt{2}}\times\dfrac{\sqrt{5}+\sqrt{2}}{\sqrt{5}+\sqrt{2}}\\ \ =\dfrac{\sqrt{5}(\sqrt{5}-\sqrt{2})}{(\sqrt{5})^2-(\sqrt{2})^2}\\ \ = \dfrac{5-\sqrt{10}}{3}\\ ( 3 )\ \dfrac{2+\sqrt{3}}{2-\sqrt{3}}\\ \ =\dfrac{2+\sqrt{3}}{2-\sqrt{3}}\times\dfrac{2+\sqrt{3}}{2+\sqrt{3}}\\ \ =\dfrac{(2+\sqrt{3})^2}{2^2-(\sqrt{3})^2}\\ \ =4+3+4\sqrt{3}\\ \ =7+4\sqrt{3}\\ ( 4 )\ \dfrac{1}{3-\sqrt{2}+\sqrt{7}}\\ \ =\dfrac{1}{3-(\sqrt{2}-\sqrt{7})}\times\dfrac{3+(\sqrt{2}-\sqrt{7})}{3+(\sqrt{2}-\sqrt{7})}\\ \ =\dfrac{3+(\sqrt{2}-\sqrt{7})}{3^2-(\sqrt{2}-\sqrt{7})^2}\\ \ =\dfrac{3+(\sqrt{2}-\sqrt{7})}{9-(9-2\sqrt{14})}\\ \ =\dfrac{3+(\sqrt{2}-\sqrt{7})}{2\sqrt{14}}\times\dfrac{\sqrt{14}}{\sqrt{14}}\\ \ =\dfrac{3\sqrt{14} +2\sqrt{7} -7\sqrt{2}}{28}\\ }$

${\fbox{4} \ \\ ( 1 )\ \sqrt{6+2\sqrt{5}}\\ \ =\sqrt{5+1+2\sqrt{5}}\\ \ =\sqrt{(\sqrt{5}+1)^2}\\ \ =\sqrt{5}+1\\ ( 2 )\ \sqrt{8-2\sqrt{12}}\\ \ =\sqrt{6+2 -2\sqrt{12}}\\ \ =\sqrt{(\sqrt{6}-\sqrt{2})^2}\\ \ =\sqrt{6}-\sqrt{2}\ (\because \sqrt{6}\gt \sqrt{2})\\ ( 3 )\ \sqrt{6+\sqrt{32}}\\ \ = \sqrt{4+2+2\sqrt{8}}\\ \ =\sqrt{(2+\sqrt{2})^2}\\ \ =2+\sqrt{2}\\ ( 4 )\ \sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\\ \ =\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}\times\dfrac{2-\sqrt{3}}{2-\sqrt{3}}}\\ \ = \sqrt{\dfrac{(2-\sqrt{3})^2}{2^2-(\sqrt{3})^2}}\\ \ =2-\sqrt{3}\ (\because 2\gt \sqrt{3})\\ }$