与えられた式 $\sqrt{6 \times \sqrt[3]{12} \times \sqrt[6]{\frac{2}{3}}}$ を計算して簡単にせよ。代数学根号指数法則式の計算2025/6/191. 問題の内容与えられた式 6×123×236\sqrt{6 \times \sqrt[3]{12} \times \sqrt[6]{\frac{2}{3}}}6×312×632 を計算して簡単にせよ。2. 解き方の手順まず、それぞれの根号を指数表現に変換します。6=612\sqrt{6} = 6^{\frac{1}{2}}6=621123=1213\sqrt[3]{12} = 12^{\frac{1}{3}}312=1231236=(23)16\sqrt[6]{\frac{2}{3}} = (\frac{2}{3})^{\frac{1}{6}}632=(32)61次に、与えられた式全体を指数表現に変換します。6×123×236=(6×1213×(23)16)12\sqrt{6 \times \sqrt[3]{12} \times \sqrt[6]{\frac{2}{3}}} = (6 \times 12^{\frac{1}{3}} \times (\frac{2}{3})^{\frac{1}{6}})^{\frac{1}{2}}6×312×632=(6×1231×(32)61)21指数法則を使って、全体を 12\frac{1}{2}21 乗します。(6×1213×(23)16)12=612×1213×12×(23)16×12=612×1216×(23)112(6 \times 12^{\frac{1}{3}} \times (\frac{2}{3})^{\frac{1}{6}})^{\frac{1}{2}} = 6^{\frac{1}{2}} \times 12^{\frac{1}{3} \times \frac{1}{2}} \times (\frac{2}{3})^{\frac{1}{6} \times \frac{1}{2}} = 6^{\frac{1}{2}} \times 12^{\frac{1}{6}} \times (\frac{2}{3})^{\frac{1}{12}}(6×1231×(32)61)21=621×1231×21×(32)61×21=621×1261×(32)12112=22×312 = 2^2 \times 312=22×3 であるから、1216=(22×3)16=226×316=213×31612^{\frac{1}{6}} = (2^2 \times 3)^{\frac{1}{6}} = 2^{\frac{2}{6}} \times 3^{\frac{1}{6}} = 2^{\frac{1}{3}} \times 3^{\frac{1}{6}}1261=(22×3)61=262×361=231×361(23)112=21123112(\frac{2}{3})^{\frac{1}{12}} = \frac{2^{\frac{1}{12}}}{3^{\frac{1}{12}}}(32)121=31212121したがって、612×1216×(23)112=(2×3)12×213×316×21123112=212×312×213×316×2112×3−1126^{\frac{1}{2}} \times 12^{\frac{1}{6}} \times (\frac{2}{3})^{\frac{1}{12}} = (2 \times 3)^{\frac{1}{2}} \times 2^{\frac{1}{3}} \times 3^{\frac{1}{6}} \times \frac{2^{\frac{1}{12}}}{3^{\frac{1}{12}}} = 2^{\frac{1}{2}} \times 3^{\frac{1}{2}} \times 2^{\frac{1}{3}} \times 3^{\frac{1}{6}} \times 2^{\frac{1}{12}} \times 3^{-\frac{1}{12}}621×1261×(32)121=(2×3)21×231×361×31212121=221×321×231×361×2121×3−121指数の和を取ります。212+13+112×312+16−112=26+4+112×36+2−112=21112×37122^{\frac{1}{2} + \frac{1}{3} + \frac{1}{12}} \times 3^{\frac{1}{2} + \frac{1}{6} - \frac{1}{12}} = 2^{\frac{6+4+1}{12}} \times 3^{\frac{6+2-1}{12}} = 2^{\frac{11}{12}} \times 3^{\frac{7}{12}}221+31+121×321+61−121=2126+4+1×3126+2−1=21211×312721112×3712=21112×3712=(211×37)112=(2048×2187)112=(4478976)1122^{\frac{11}{12}} \times 3^{\frac{7}{12}} = 2^{\frac{11}{12}} \times 3^{\frac{7}{12}} = (2^{11} \times 3^7)^{\frac{1}{12}} = (2048 \times 2187)^{\frac{1}{12}} = (4478976)^{\frac{1}{12}}21211×3127=21211×3127=(211×37)121=(2048×2187)121=(4478976)121この形では簡単にならないようです。計算を見直します。612×1216×(23)112=612×1216×2112×3−112=6×126×2312=(2⋅3)12(22⋅3)16(23)112=21231221331621123−112=261236122412321221123−112=2111237126^{\frac{1}{2}} \times 12^{\frac{1}{6}} \times (\frac{2}{3})^{\frac{1}{12}} = 6^{\frac{1}{2}} \times 12^{\frac{1}{6}} \times 2^{\frac{1}{12}} \times 3^{-\frac{1}{12}} = \sqrt{6} \times \sqrt[6]{12} \times \sqrt[12]{\frac{2}{3}} = (2 \cdot 3)^{\frac{1}{2}}(2^2 \cdot 3)^{\frac{1}{6}}(\frac{2}{3})^{\frac{1}{12}}= 2^{\frac{1}{2}}3^{\frac{1}{2}}2^{\frac{1}{3}}3^{\frac{1}{6}}2^{\frac{1}{12}}3^{-\frac{1}{12}}= 2^{\frac{6}{12}}3^{\frac{6}{12}}2^{\frac{4}{12}}3^{\frac{2}{12}}2^{\frac{1}{12}}3^{-\frac{1}{12}} = 2^{\frac{11}{12}}3^{\frac{7}{12}}621×1261×(32)121=621×1261×2121×3−121=6×612×1232=(2⋅3)21(22⋅3)61(32)121=22132123136121213−121=212631262124312221213−121=212113127211123712=(21137)112=(21137)112=21137122^{\frac{11}{12}}3^{\frac{7}{12}}= (2^{11}3^7)^{\frac{1}{12}}= (2^{11}3^7)^{\frac{1}{12}} = \sqrt[12]{2^{11} 3^7}212113127=(21137)121=(21137)121=12211373. 最終的な答え2113712\sqrt[12]{2^{11}3^7}1221137