問題45から47は、それぞれ根号や指数を含む式の値を求める問題です。 45: (1) $\sqrt[4]{16}$ (2) $-\sqrt[3]{64}$ 46: (1) $\sqrt[4]{4} \times \sqrt[3]{16}$ (2) $\frac{\sqrt[3]{48}}{\sqrt[3]{3}}$ (3) $(\sqrt[4]{5})^8$ (4) $\sqrt{\sqrt[3]{729}}$ 47: (1) $8^{\frac{1}{3}}$ (2) $125^{\frac{2}{3}}$ (3) $4^{-\frac{1}{2}}$ (4) $0.04^{1.5}$ (5) $(\frac{27}{8})^{-\frac{4}{3}}$

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1. 問題の内容

問題45から47は、それぞれ根号や指数を含む式の値を求める問題です。

45: (1) $\sqrt[4]{16}$ (2) $-\sqrt[3]{64}$

46: (1) $\sqrt[4]{4} \times \sqrt[3]{16}$ (2) $\frac{\sqrt[3]{48}}{\sqrt[3]{3}}$ (3) $(\sqrt[4]{5})^8$ (4) $\sqrt{\sqrt[3]{729}}$

47: (1) $8^{\frac{1}{3}}$ (2) $125^{\frac{2}{3}}$ (3) $4^{-\frac{1}{2}}$ (4) $0.04^{1.5}$ (5) $(\frac{27}{8})^{-\frac{4}{3}}$

2. 解き方の手順

45:

(1) $\sqrt[4]{16} = \sqrt[4]{2^4} = 2$

(2) $-\sqrt[3]{64} = -\sqrt[3]{4^3} = -4$

46:

(1) $\sqrt[4]{4} \times \sqrt[3]{16} = (2^2)^{\frac{1}{4}} \times (2^4)^{\frac{1}{3}} = 2^{\frac{1}{2}} \times 2^{\frac{4}{3}} = 2^{\frac{1}{2} + \frac{4}{3}} = 2^{\frac{3+8}{6}} = 2^{\frac{11}{6}} = \sqrt[6]{2^{11}} = \sqrt[6]{2048}$

(2) $\frac{\sqrt[3]{48}}{\sqrt[3]{3}} = \sqrt[3]{\frac{48}{3}} = \sqrt[3]{16} = \sqrt[3]{2^4} = 2\sqrt[3]{2}$

(3) $(\sqrt[4]{5})^8 = (5^{\frac{1}{4}})^8 = 5^{\frac{8}{4}} = 5^2 = 25$

(4) $\sqrt{\sqrt[3]{729}} = \sqrt[6]{729} = \sqrt[6]{3^6} = 3$

47:

(1) $8^{\frac{1}{3}} = (2^3)^{\frac{1}{3}} = 2$

(2) $125^{\frac{2}{3}} = (5^3)^{\frac{2}{3}} = 5^2 = 25$

(3) $4^{-\frac{1}{2}} = (2^2)^{-\frac{1}{2}} = 2^{-1} = \frac{1}{2}$

(4) $0.04^{1.5} = (\frac{4}{100})^{\frac{3}{2}} = (\frac{1}{25})^{\frac{3}{2}} = ((\frac{1}{5})^2)^{\frac{3}{2}} = (\frac{1}{5})^3 = \frac{1}{125} = 0.008$

(5) $(\frac{27}{8})^{-\frac{4}{3}} = (\frac{3^3}{2^3})^{-\frac{4}{3}} = ((\frac{3}{2})^3)^{-\frac{4}{3}} = (\frac{3}{2})^{-4} = (\frac{2}{3})^4 = \frac{16}{81}$

3. 最終的な答え

45: (1) 2 (2) -4

46: (1) $\sqrt[6]{2048}$ (2) $2\sqrt[3]{2}$ (3) 25 (4) 3

47: (1) 2 (2) 25 (3) $\frac{1}{2}$ (4) $\frac{1}{125}$ (5) $\frac{16}{81}$