【高校数学Ⅰ】1-2-3 平方根の応用｜問題集
1.\(\displaystyle x=\frac{\sqrt{3}+1}{\sqrt{3}-1},y=2-\sqrt{3}\)のとき、次の式の値を求めなさい。
(1)\(x+y\)
\(x\)を有理化する。
\(\displaystyle x=\frac{\sqrt{3}+1}{\sqrt{3}-1}\)
\(\displaystyle \ \ =\frac{(\sqrt{3}+1)^2}{(\sqrt{3}-1)(\sqrt{3}+1)}\)
\(\displaystyle \ \ =\frac{3+2\sqrt{3}+1}{3-1}\)
\(\displaystyle \ \ =\frac{4+2\sqrt{3}}{2}\)
\(\displaystyle \ \ =2+\sqrt{3}\)
よって、
\(x+y\)
\(=(2+\sqrt{3})+(2-\sqrt{3})\)
\(=2+\sqrt{3}+2-\sqrt{3}\)
\(=4\)
\(\displaystyle x=\frac{\sqrt{3}+1}{\sqrt{3}-1}\)
\(\displaystyle \ \ =\frac{(\sqrt{3}+1)^2}{(\sqrt{3}-1)(\sqrt{3}+1)}\)
\(\displaystyle \ \ =\frac{3+2\sqrt{3}+1}{3-1}\)
\(\displaystyle \ \ =\frac{4+2\sqrt{3}}{2}\)
\(\displaystyle \ \ =2+\sqrt{3}\)
よって、
\(x+y\)
\(=(2+\sqrt{3})+(2-\sqrt{3})\)
\(=2+\sqrt{3}+2-\sqrt{3}\)
\(=4\)
(2)\(xy\)
\(=(2+\sqrt{3})(2-\sqrt{3})\)
\(=4-3\)
\(=1\)
\(=4-3\)
\(=1\)
(3)\(x^2+y^2\)
\(=(x+y)^2-2xy\)
\(=4^2-2×1\)
\(=16-2\)
\(=14\)
\(=4^2-2×1\)
\(=16-2\)
\(=14\)
2.\(\displaystyle x=\frac{2-\sqrt{2}}{2+\sqrt{2}},y=\frac{2+\sqrt{2}}{2-\sqrt{2}}\)のとき、次の式の値を求めなさい。
(1)\(x+y\)
\(x,y\)をそれぞれ有理化する。
\(\displaystyle x=\frac{2-\sqrt{2}}{2+\sqrt{2}}\)
\(\displaystyle \ \ =\frac{(2-\sqrt{2})^2}{(2-\sqrt{2})(2+\sqrt{2})}\)
\(\displaystyle \ \ =\frac{4-4\sqrt{2}+2}{4-2}\)
\(\displaystyle \ \ =\frac{6-4\sqrt{2}}{2}\)
\(\displaystyle \ \ =3-2\sqrt{2}\)
\(\displaystyle y=\frac{2+\sqrt{2}}{2-\sqrt{2}}\)
\(\displaystyle \ \ =\frac{(2+\sqrt{2})^2}{(2-\sqrt{2})(2+\sqrt{2})}\)
\(\displaystyle \ \ =\frac{4+4\sqrt{2}+2}{4-2}\)
\(\displaystyle \ \ =\frac{6+4\sqrt{2}}{2}\)
\(\displaystyle \ \ =3+2\sqrt{2}\)
よって、
\(x+y\)
\(=(3-2\sqrt{2})+(3+2\sqrt{2})\)
\(=6\)
\(\displaystyle x=\frac{2-\sqrt{2}}{2+\sqrt{2}}\)
\(\displaystyle \ \ =\frac{(2-\sqrt{2})^2}{(2-\sqrt{2})(2+\sqrt{2})}\)
\(\displaystyle \ \ =\frac{4-4\sqrt{2}+2}{4-2}\)
\(\displaystyle \ \ =\frac{6-4\sqrt{2}}{2}\)
\(\displaystyle \ \ =3-2\sqrt{2}\)
\(\displaystyle y=\frac{2+\sqrt{2}}{2-\sqrt{2}}\)
\(\displaystyle \ \ =\frac{(2+\sqrt{2})^2}{(2-\sqrt{2})(2+\sqrt{2})}\)
\(\displaystyle \ \ =\frac{4+4\sqrt{2}+2}{4-2}\)
\(\displaystyle \ \ =\frac{6+4\sqrt{2}}{2}\)
\(\displaystyle \ \ =3+2\sqrt{2}\)
よって、
\(x+y\)
\(=(3-2\sqrt{2})+(3+2\sqrt{2})\)
\(=6\)
(2)\(xy\)
\(=(3-2\sqrt{2})(3+2\sqrt{2})\)
\(=9-8\)
\(=1\)
\(=9-8\)
\(=1\)
(3)\(x^2+y^2\)
\(=(x+y)^2-2xy\)
\(=6^2-2×1\)
\(=36-2\)
\(=34\)
\(=6^2-2×1\)
\(=36-2\)
\(=34\)
(4)\(x^3+y^3\)
\(=(x+y)^3-3xy(x+y)\)
\(=6^3-3×1×6\)
\(=216-18\)
\(=198\)
\(=6^3-3×1×6\)
\(=216-18\)
\(=198\)
3.\(\displaystyle x=\frac{3-\sqrt{6}}{3+\sqrt{6}},y=\frac{3+\sqrt{6}}{3-\sqrt{6}}\)のとき、次の式の値を求めなさい。
(1)\(x+y\)
\(x,y\)をそれぞれ有理化する。
\(\displaystyle x=\frac{3-\sqrt{6}}{3+\sqrt{6}}\)
\(\displaystyle \ \ =\frac{(3-\sqrt{6})^2}{(3-\sqrt{6})(3+\sqrt{6})}\)
\(\displaystyle \ \ =\frac{9-6\sqrt{6}+6}{9-6}\)
\(\displaystyle \ \ =\frac{15-6\sqrt{6}}{3}\)
\(\displaystyle \ \ =5-2\sqrt{6}\)
\(\displaystyle y=\frac{3+\sqrt{6}}{3-\sqrt{6}}\)
\(\displaystyle \ \ =\frac{(3+\sqrt{6})^2}{(3-\sqrt{6})(3+\sqrt{6})}\)
\(\displaystyle \ \ =\frac{9+6\sqrt{6}+6}{9-6}\)
\(\displaystyle \ \ =\frac{15+6\sqrt{6}}{3}\)
\(\displaystyle \ \ =5+2\sqrt{6}\)
よって、
\(x+y\)
\(=(5-2\sqrt{6})+(5+2\sqrt{6})\)
\(=10\)
\(\displaystyle x=\frac{3-\sqrt{6}}{3+\sqrt{6}}\)
\(\displaystyle \ \ =\frac{(3-\sqrt{6})^2}{(3-\sqrt{6})(3+\sqrt{6})}\)
\(\displaystyle \ \ =\frac{9-6\sqrt{6}+6}{9-6}\)
\(\displaystyle \ \ =\frac{15-6\sqrt{6}}{3}\)
\(\displaystyle \ \ =5-2\sqrt{6}\)
\(\displaystyle y=\frac{3+\sqrt{6}}{3-\sqrt{6}}\)
\(\displaystyle \ \ =\frac{(3+\sqrt{6})^2}{(3-\sqrt{6})(3+\sqrt{6})}\)
\(\displaystyle \ \ =\frac{9+6\sqrt{6}+6}{9-6}\)
\(\displaystyle \ \ =\frac{15+6\sqrt{6}}{3}\)
\(\displaystyle \ \ =5+2\sqrt{6}\)
よって、
\(x+y\)
\(=(5-2\sqrt{6})+(5+2\sqrt{6})\)
\(=10\)
(2)\(xy\)
\(=(5-2\sqrt{6})(5+2\sqrt{6})\)
\(=25-24\)
\(=1\)
\(=25-24\)
\(=1\)
(3)\(x^2+y^2\)
\(=(x+y)^2-2xy\)
\(=10^2-2×1\)
\(=100-2\)
\(=98\)
\(=10^2-2×1\)
\(=100-2\)
\(=98\)
(4)\(x^3+y^3\)
\(=(x+y)^3-3xy(x+y)\)
\(=10^3-3×1×10\)
\(=1000-30\)
\(=970\)
\(=10^3-3×1×10\)
\(=1000-30\)
\(=970\)
4.\(\displaystyle \frac{1}{2-\sqrt{3}}\)の整数部分を\(a\)、小数部分を\(b\)としたとき、次の式の値を求めなさい。
(1)\(a\)
有理化する。
\(\displaystyle =\frac{2+\sqrt{3}}{(2-\sqrt{3})(2+\sqrt{3})}\)
\(\displaystyle =\frac{2+\sqrt{3}}{4-3}\)
\(=2+\sqrt{3}\)
\(1<\sqrt{3}<2\)より、
\(3<2+\sqrt{3}<4\)
よって、整数部分は\(3\)となる。
\(a=3\)
\(\displaystyle =\frac{2+\sqrt{3}}{(2-\sqrt{3})(2+\sqrt{3})}\)
\(\displaystyle =\frac{2+\sqrt{3}}{4-3}\)
\(=2+\sqrt{3}\)
\(1<\sqrt{3}<2\)より、
\(3<2+\sqrt{3}<4\)
よって、整数部分は\(3\)となる。
\(a=3\)
(2)\(b\)
小数部分は、\(2+\sqrt{3}-3=\sqrt{3}-1\)
\(b=\sqrt{3}-1\)
\(b=\sqrt{3}-1\)
(3)\(a+2b+b^2\)
\(=3+2(\sqrt{3}-1)+(\sqrt{3}-1)^2\)
\(=3+2\sqrt{3}-2+3-2\sqrt{3}+1\)
\(=5\)
\(=3+2\sqrt{3}-2+3-2\sqrt{3}+1\)
\(=5\)
5.\(\sqrt{6}\)の整数部分を\(a\)、小数部分を\(b\)としたとき、次の式の値を求めなさい。
(1)\(a\)
\(2<\sqrt{6}<3\)より、
よって、整数部分は\(2\)となる。
\(a=2\)
よって、整数部分は\(2\)となる。
\(a=2\)
(2)\(b\)
小数部分は、\(\sqrt{6}-2\)
\(b=\sqrt{6}-2\)
\(b=\sqrt{6}-2\)
(3)\(\displaystyle \frac{a}{b}\)
\(\displaystyle =\frac{2}{\sqrt{6}-2}\)
\(\displaystyle =\frac{2(\sqrt{6}+2)}{(\sqrt{6}-2)(\sqrt{6}+2)}\)
\(\displaystyle =\frac{2\sqrt{6}+4}{6-4}\)
\(\displaystyle =\frac{2\sqrt{6}+4}{2}\)
\(=\sqrt{6}+2\)
\(\displaystyle =\frac{2(\sqrt{6}+2)}{(\sqrt{6}-2)(\sqrt{6}+2)}\)
\(\displaystyle =\frac{2\sqrt{6}+4}{6-4}\)
\(\displaystyle =\frac{2\sqrt{6}+4}{2}\)
\(=\sqrt{6}+2\)
6.\(2\sqrt{3}\)の整数部分を\(a\)、小数部分を\(b\)としたとき、次の式の値を求めなさい。
(1)\(a\)
\(2\sqrt{3}=\sqrt{12}\)
\(3<\sqrt{12}<4\)より、
よって、整数部分は\(3\)となる。
\(a=3\)
\(3<\sqrt{12}<4\)より、
よって、整数部分は\(3\)となる。
\(a=3\)
(2)\(b\)
小数部分は、\(2\sqrt{3}-3\)
\(b=2\sqrt{3}-3\)
\(b=2\sqrt{3}-3\)
(3)\(\displaystyle \frac{a}{b}\)
\(\displaystyle =\frac{3}{2\sqrt{3}-3}\)
\(\displaystyle =\frac{3(2\sqrt{3}+3)}{(2\sqrt{3}-3)(2\sqrt{3}+3)}\)
\(\displaystyle =\frac{6\sqrt{3}+9}{12-9}\)
\(\displaystyle =\frac{6\sqrt{3}+9}{3}\)
\(=2\sqrt{3}+3\)
\(\displaystyle =\frac{3(2\sqrt{3}+3)}{(2\sqrt{3}-3)(2\sqrt{3}+3)}\)
\(\displaystyle =\frac{6\sqrt{3}+9}{12-9}\)
\(\displaystyle =\frac{6\sqrt{3}+9}{3}\)
\(=2\sqrt{3}+3\)
7.次の式を簡単にしなさい。
(1)\(\sqrt{9-2\sqrt{14}}\)
\(=\sqrt{7}-\sqrt{2}\)
(2)\(\sqrt{15+6\sqrt{6}}\)
\(=\sqrt{15+2\sqrt{54}}\)
\(=\sqrt{9}+\sqrt{6}\)
\(=3+\sqrt{6}\)
\(=\sqrt{9}+\sqrt{6}\)
\(=3+\sqrt{6}\)
(3)\(\sqrt{3-\sqrt{5}}\)
\(\displaystyle =\sqrt{\frac{6-2\sqrt{5}}{2}}\)
\(\displaystyle =\frac{\sqrt{5}-\sqrt{1}}{\sqrt{2}}\)
\(\displaystyle =\frac{\sqrt{10}-\sqrt{2}}{2}\)
\(\displaystyle =\frac{\sqrt{5}-\sqrt{1}}{\sqrt{2}}\)
\(\displaystyle =\frac{\sqrt{10}-\sqrt{2}}{2}\)
(4)\(\sqrt{8+2\sqrt{15}}\)
\(=\sqrt{5}+\sqrt{3}\)
(5)\(\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{5}-\sqrt{1}\)
(6)\(\sqrt{9+4\sqrt{5}}\)
\(=\sqrt{9+2\sqrt{20}}\)
\(=\sqrt{5}+\sqrt{4}\)
\(=\sqrt{5}+2\)
\(=\sqrt{5}+\sqrt{4}\)
\(=\sqrt{5}+2\)
(7)\(\sqrt{4-\sqrt{7}}\)
\(\displaystyle =\sqrt{\frac{8-2\sqrt{7}}{2}}\)
\(\displaystyle =\frac{\sqrt{7}-\sqrt{1}}{\sqrt{2}}\)
\(\displaystyle =\frac{\sqrt{14}-\sqrt{2}}{2}\)
\(\displaystyle =\frac{\sqrt{7}-\sqrt{1}}{\sqrt{2}}\)
\(\displaystyle =\frac{\sqrt{14}-\sqrt{2}}{2}\)
(8)\(\sqrt{3+2\sqrt{2}}\)
\(=\sqrt{2}+\sqrt{1}\)
\(=\sqrt{2}+1\)
\(=\sqrt{2}+1\)
(9)\(\sqrt{7-2\sqrt{6}}\)
\(=\sqrt{6}-\sqrt{1}\)
\(=\sqrt{6}-1\)
\(=\sqrt{6}-1\)
(10)\(\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{7-2\sqrt{12}}\)
\(=\sqrt{4}-\sqrt{3}\)
\(=2-\sqrt{3}\)
\(=\sqrt{4}-\sqrt{3}\)
\(=2-\sqrt{3}\)
(11)\(\sqrt{2+\sqrt{3}}\)
\(\displaystyle =\sqrt{\frac{4+2\sqrt{3}}{2}}\)
\(\displaystyle =\frac{\sqrt{3}+\sqrt{1}}{\sqrt{2}}\)
\(\displaystyle =\frac{\sqrt{6}+\sqrt{2}}{2}\)
\(\displaystyle =\frac{\sqrt{3}+\sqrt{1}}{\sqrt{2}}\)
\(\displaystyle =\frac{\sqrt{6}+\sqrt{2}}{2}\)
(12)\(\sqrt{8-4\sqrt{3}}\)
\(=\sqrt{8-2\sqrt{12}}\)
\(=\sqrt{6}-\sqrt{2}\)
\(=\sqrt{6}-\sqrt{2}\)
(13)\(\sqrt{4+\sqrt{15}}\)
\(\displaystyle =\sqrt{\frac{8+2\sqrt{15}}{2}}\)
\(\displaystyle =\frac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}\)
\(\displaystyle =\frac{\sqrt{10}+\sqrt{6}}{2}\)
\(\displaystyle =\frac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}\)
\(\displaystyle =\frac{\sqrt{10}+\sqrt{6}}{2}\)
(14)\(\sqrt{6+2\sqrt{5}}\)
\(=\sqrt{5}+\sqrt{1}\)
\(=\sqrt{5}+1\)
\(=\sqrt{5}+1\)
(15)\(\sqrt{12-6\sqrt{3}}\)
\(=\sqrt{12-2\sqrt{27}}\)
\(=\sqrt{9}-\sqrt{3}\)
\(=3-\sqrt{3}\)
\(=\sqrt{9}-\sqrt{3}\)
\(=3-\sqrt{3}\)
(16)\(\sqrt{7+2\sqrt{10}}\)
\(=\sqrt{5}+\sqrt{2}\)
次の学習に進もう！