【高校数学Ⅱ】5-1-1 指数法則｜問題集
1.次の計算をしなさい。
(1)\(a^5×a^4\)
\(=a^9\)
(2)\(a^6÷a^3\)
\(=a^3\)
(3)\((2a^2)^4\)
\(=16a^8\)
(4)\(3^2÷3^{-1}\)
\(=3^3\)
\(=27\)
\(=27\)
(5)\(a^5×a^{-2}÷a^{-3}\)
\(=a^6\)
(6)\(3^5×9^{-4}÷27^{-2}\)
\(=3^5×3^{-8}÷3^{-6}\)
\(=3^3\)
\(=27\)
\(=3^3\)
\(=27\)
(7)\(12^3×2^{-4}\)
\(=(2^2×3)^3×2^{-4}\)
\(=2^6×3^3×2^{-4}\)
\(=2^2×3^3\)
\(=108\)
\(=2^6×3^3×2^{-4}\)
\(=2^2×3^3\)
\(=108\)
(8)\(\displaystyle \left(\frac{1}{2}\right)^2×4^3÷2^4\)
\(=2^{-2}×2^6×2^{-4}\)
\(=2^0\)
\(=1\)
\(=2^0\)
\(=1\)
(9)\(\displaystyle \left(\frac{a}{b}\right)^3÷\frac{b^3}{a^2}×\left(\frac{b}{a^2}\right)^4\)
\(=a^3b^{-3}×a^2b^{-3}×a^{-8}b^4\)
\(=a^{-3}b^{-2}\)
\(\displaystyle =\frac{1}{a^3b^2}\)
\(=a^{-3}b^{-2}\)
\(\displaystyle =\frac{1}{a^3b^2}\)
(10)\(\displaystyle 9^2×\frac{1}{27}÷3^3\)
\(=3^4×3^{-3}×3^{-3}\)
\(=3^{-2}\)
\(\displaystyle =\frac{1}{9}\)
\(=3^{-2}\)
\(\displaystyle =\frac{1}{9}\)
(11)\(\sqrt[4]{16}\)
\(=\sqrt[4]{2^4}\)
\(=2\)
\(=2\)
(12)\(\sqrt[3]{27}\)
\(=\sqrt[3]{3^3}\)
\(=3\)
\(=3\)
(13)\(\sqrt[3]{-27}\)
\(=\sqrt[3]{(-3)^3}\)
\(=-3\)
\(=-3\)
(14)\(\sqrt[3]{25}×\sqrt[3]{5}\)
\(=\sqrt[3]{5^3}\)
\(=5\)
\(=5\)
(15)\((\sqrt[3]{3})^9\)
\(=3^3\)
\(=27\)
\(=27\)
(16)\(\displaystyle \frac{\sqrt[3]{48}}{\sqrt[3]{6}}\)
\(=\sqrt[3]{8}\)
\(=\sqrt[3]{2^3}\)
\(=2\)
\(=\sqrt[3]{2^3}\)
\(=2\)
(17)\(\sqrt[4]{\sqrt{256}}\)
\(=\sqrt[8]{2^8}\)
\(=2\)
\(=2\)
(18)\(8^{\frac{2}{3}}×4^{\frac{3}{2}}\)
\(=2^2×2^3\)
\(=2^5\)
\(=32\)
\(=2^5\)
\(=32\)
(19)\(2^{-\frac{1}{2}}×2^{\frac{5}{6}}÷2^{\frac{1}{3}}\)
\(=2^0\)
\(=1\)
\(=1\)
(20)\((3^{-2}×9^{\frac{2}{3}})^{\frac{3}{2}}\)
\(=(3^{-2}×3^{\frac{4}{3}})^{\frac{3}{2}}\)
\(=(3^{-\frac{2}{3}})^{\frac{3}{2}}\)
\(=3^{-1}\)
\(\displaystyle =\frac{1}{3}\)
\(=(3^{-\frac{2}{3}})^{\frac{3}{2}}\)
\(=3^{-1}\)
\(\displaystyle =\frac{1}{3}\)
(21)\(\sqrt[3]{5}÷\sqrt[12]{5}×\sqrt[8]{25}\)
\(=5^{\frac{1}{3}}÷5^{\frac{1}{12}}×5^{\frac{1}{4}}\)
\(=5^{\frac{1}{2}}\)
\(=\sqrt{5}\)
\(=5^{\frac{1}{2}}\)
\(=\sqrt{5}\)
(22)\(\sqrt{6}×\sqrt[4]{54}÷\sqrt[4]{6}\)
\(=(2×3)^{\frac{1}{2}}×(2×3^3)^{\frac{1}{4}}×(2×3)^{-\frac{1}{4}}\)
\(=2^{\frac{1}{2}}×3^1\)
\(=3\sqrt{2}\)
\(=2^{\frac{1}{2}}×3^1\)
\(=3\sqrt{2}\)
2.\(x^{\frac{1}{2}}+x^{-\frac{1}{2}}=3\)のとき、次の値を求めなさい。ただし、\(x>0\)とする。
(1)\(x+x^{-1}\)
\(x^{\frac{1}{2}}+x^{-\frac{1}{2}}=3\)の両辺を\(2\)乗すると、
\(x+2+x^{-1}=9\)
\(x+x^{-1}=7\)
\(x+2+x^{-1}=9\)
\(x+x^{-1}=7\)
(2)\(x^{\frac{3}{2}}+x^{-\frac{3}{2}}\)
\(=(x^{\frac{1}{2}})^3+(x^{-\frac{1}{2}})^3\)
\(=(x^{\frac{1}{2}}+x^{-\frac{1}{2}})\{(x^{\frac{1}{2}})^2-x^{\frac{1}{2}}x^{-\frac{1}{2}}+(x^{-\frac{1}{2}})^2\}\)
\(=(x^{\frac{1}{2}}+x^{-\frac{1}{2}})(x-1+x^{-1})\)
\(=3(7-1)\)
\(=18\)
\(=(x^{\frac{1}{2}}+x^{-\frac{1}{2}})\{(x^{\frac{1}{2}})^2-x^{\frac{1}{2}}x^{-\frac{1}{2}}+(x^{-\frac{1}{2}})^2\}\)
\(=(x^{\frac{1}{2}}+x^{-\frac{1}{2}})(x-1+x^{-1})\)
\(=3(7-1)\)
\(=18\)
3.\(2^x-2^{-x}=3\)のとき、次の値を求めなさい。
(1)\(4^x+4^{-x}\)
\(2^x-2^{-x}=3\)の両辺を\(2\)乗すると、
\(2^{2x}-2+2^{-2x}=9\)
\(4^x+4^{-x}=11\)
\(2^{2x}-2+2^{-2x}=9\)
\(4^x+4^{-x}=11\)
(2)\(2^x+2^{-x}\)
\((2^x+2^{-x})^2\)
\(=2^{2x}+2+2^{-2x}\)
\(=4^{x}+4^{-x}+2\)
\(=11+2\)
\(=13\)
\(2^x+2^{-x}>0\)より、
\(2^x+2^{-x}=\sqrt{13}\)
\(=2^{2x}+2+2^{-2x}\)
\(=4^{x}+4^{-x}+2\)
\(=11+2\)
\(=13\)
\(2^x+2^{-x}>0\)より、
\(2^x+2^{-x}=\sqrt{13}\)
次の学習に進もう！