【高校数学Ⅰ】1-2-1 実数｜問題集
1.次の分数を循環小数で表しなさい。
(1)\(\displaystyle \frac{1}{3}\)
\(=0.\dot{3}\)
(2)\(\displaystyle \frac{2}{11}\)
\(=0.\dot{1}\dot{8}\)
(3)\(\displaystyle \frac{31}{27}\)
\(=1.\dot{1}4\dot{8}\)
(4)\(\displaystyle \frac{1}{6}\)
\(=0.1\dot{6}\)
(5)\(\displaystyle \frac{1}{11}\)
\(=0.\dot{0}\dot{9}\)
(6)\(\displaystyle \frac{1}{12}\)
\(=0.08\dot{3}\)
(7)\(\displaystyle \frac{3}{7}\)
\(=0.\dot{4}2857\dot{1}\)
2.次の循環小数を分数で表しなさい。
(1)\(0.\dot{1}\)
\(x=0.\dot{1}\)とおくと、\(10x=1.\dot{1}\)
\(10x-x=1.\dot{1}-0.\dot{1}\)
\(9x=1\)
\(\displaystyle x=\frac{1}{9}\)
\(10x-x=1.\dot{1}-0.\dot{1}\)
\(9x=1\)
\(\displaystyle x=\frac{1}{9}\)
(2)\(0.\dot{2}\dot{7}\)
\(x=0.\dot{2}\dot{7}\)とおくと、\(100x=27.\dot{2}\dot{7}\)
\(100x-x=27.\dot{2}\dot{7}-0.27\dot{2}\dot{7}\)
\(99x=27\)
\(\displaystyle x=\frac{3}{11}\)
\(100x-x=27.\dot{2}\dot{7}-0.27\dot{2}\dot{7}\)
\(99x=27\)
\(\displaystyle x=\frac{3}{11}\)
(3)\(0.\dot{6}4\dot{8}\)
\(x=0.\dot{6}4\dot{8}\)とおくと、\(1000x=648.\dot{6}4\dot{8}\)
\(1000x-x=648.\dot{6}4\dot{8}-0.\dot{6}4\dot{8}\)
\(999x=648\)
\(\displaystyle x=\frac{24}{37}\)
\(1000x-x=648.\dot{6}4\dot{8}-0.\dot{6}4\dot{8}\)
\(999x=648\)
\(\displaystyle x=\frac{24}{37}\)
(4)\(0.2\dot{5}\dot{4}\)
\(x=0.2\dot{5}\dot{4}\)とおくと、\(10x=2.\dot{5}\dot{4}\)、\(1000x=254.\dot{5}\dot{4}\)
\(1000x-10x=254.\dot{5}\dot{4}-2.\dot{5}\dot{4}\)
\(990x=252\)
\(\displaystyle x=\frac{14}{55}\)
\(1000x-10x=254.\dot{5}\dot{4}-2.\dot{5}\dot{4}\)
\(990x=252\)
\(\displaystyle x=\frac{14}{55}\)
(5)\(0.\dot{5}\)
\(x=0.\dot{5}\)とおくと、\(10x=5.\dot{5}\)
\(10x-x=5.\dot{5}-0.\dot{5}\)
\(9x=5\)
\(\displaystyle x=\frac{5}{9}\)
\(10x-x=5.\dot{5}-0.\dot{5}\)
\(9x=5\)
\(\displaystyle x=\frac{5}{9}\)
(6)\(0.\dot{0}\dot{7}\)
\(x=0.\dot{0}\dot{7}\)とおくと、\(100x=7.\dot{0}\dot{7}\)
\(100x-x=7.\dot{0}\dot{7}-0.\dot{0}\dot{7}\)
\(99x=7\)
\(\displaystyle x=\frac{7}{99}\)
\(100x-x=7.\dot{0}\dot{7}-0.\dot{0}\dot{7}\)
\(99x=7\)
\(\displaystyle x=\frac{7}{99}\)
(7)\(1.\dot{2}3\dot{4}\)
\(x=1.\dot{2}3\dot{4}\)とおくと、\(1000x=1234.\dot{2}3\dot{4}\)
\(1000x-x=1234.\dot{2}3\dot{4}-1.\dot{2}3\dot{4}\)
\(999x=1233\)
\(\displaystyle x=\frac{137}{111}\)
\(1000x-x=1234.\dot{2}3\dot{4}-1.\dot{2}3\dot{4}\)
\(999x=1233\)
\(\displaystyle x=\frac{137}{111}\)
3.次の値を求めなさい。
(1)\(|-6|\)
\(=6\)
(2)\(|5-8|\)
\(=|-3|\)
\(=3\)
\(=3\)
(3)\(|5|-|8|\)
\(=5-8\)
\(=-3\)
\(=-3\)
(4)\(|2-\sqrt{5}|\)
\(2-\sqrt{5}<0\)なので、
\(=\sqrt{5}-2\)
\(=\sqrt{5}-2\)
(5)\(|-3|\)
\(=3\)
(6)\(|\sqrt{5}-3|\)
\(\sqrt{5}-3<0\)なので、
\(=3-\sqrt{5}\)
\(=3-\sqrt{5}\)
(7)\(|\sqrt{3}-2||\sqrt{3}+2|\)
\(|(\sqrt{3}-2)(\sqrt{3}+2)|\)
\(=|3-4|\)
\(=1\)
\(=|3-4|\)
\(=1\)
(8)\(\displaystyle \left|-\frac{1}{2}\right|\)
\(\displaystyle =\frac{1}{2}\)
(9)\(|\sqrt{2}-\sqrt{3}|\)
\(\sqrt{2}-\sqrt{3}<0\)なので、
\(=\sqrt{3}-\sqrt{2}\)
\(=\sqrt{3}-\sqrt{2}\)
(10)\(|1|-|-2|\)
\(=1-2\)
\(=-1\)
\(=-1\)
(11)\(|2+\sqrt{5}||2-\sqrt{5}|\)
\(=|(2+\sqrt{5})(2-\sqrt{5})|\)
\(=|4-5|\)
\(=1\)
\(=|4-5|\)
\(=1\)
(12)\(|2|\)
\(=2\)
(13)\(|-11|\)
\(=11\)
(14)\(\displaystyle \left|-\frac{3}{5}\right|\)
\(\displaystyle =\frac{3}{5}\)
(15)\(|1-\sqrt{2}|\)
\(1-\sqrt{2}<0\)なので、
\(=\sqrt{2}-1\)
\(=\sqrt{2}-1\)
(16)\(|-1|\)
\(=1\)
(17)\(|15|\)
\(=15\)
(18)\(\displaystyle \left|\frac{1}{5}\right|\)
\(\displaystyle =\frac{1}{5}\)
(19)\(|2-\sqrt{5}|\)
\(2-\sqrt{5}<0\)なので、
\(=\sqrt{5}-2\)
\(=\sqrt{5}-2\)
次の学習に進もう！