3乗根の有理化
3乗根の有理化
次の式を有理化せよ。
\[ \frac{1}{2\cdot3^{\frac{2}{3}}+3\cdot3^{\frac{1}{3}}+2} \]
次の式を有理化せよ。
\[ \frac{1}{2\cdot3^{\frac{2}{3}}+3\cdot3^{\frac{1}{3}}+2} \]
(0)
\begin{align*} \frac{1}{2\cdot3^{\frac{2}{3}}+3\cdot3^{\frac{1}{3}}+2} & =\frac{\left(2\cdot3^{\frac{1}{3}}-3\right)}{\left(2\cdot3^{\frac{2}{3}}+3\cdot3^{\frac{1}{3}}+2\right)\left(2\cdot3^{\frac{1}{3}}-3\right)}\\ & =\frac{2\cdot3^{\frac{1}{3}}-3}{2^{2}\cdot3+\left(-3^{2}+2^{2}\right)\cdot3^{\frac{1}{3}}-6}\\ & =\frac{2\cdot3^{\frac{1}{3}}-3}{-5\cdot3^{\frac{1}{3}}+6}\\ & =\frac{\left(2\cdot3^{\frac{1}{3}}-3\right)\left(\left(-5\cdot3^{\frac{1}{3}}\right)^{2}+5\cdot3^{\frac{1}{3}}\cdot6+6^{2}\right)}{\left(-5\cdot3^{\frac{1}{3}}+6\right)\left(\left(-5\cdot3^{\frac{1}{3}}\right)^{2}+5\cdot3^{\frac{1}{3}}\cdot6+6^{2}\right)}\\ & =\frac{\left(2\cdot3^{\frac{1}{3}}-3\right)\left(\left(-5\cdot3^{\frac{1}{3}}\right)^{2}+5\cdot3^{\frac{1}{3}}\cdot6+6^{2}\right)}{\left(-5\cdot3^{\frac{1}{3}}\right)^{3}+6^{3}}\\ & =\frac{\left(2\cdot3^{\frac{1}{3}}-3\right)\left(25\cdot3^{\frac{2}{3}}+30\cdot3^{\frac{1}{3}}+36\right)}{-159}\\ & =\frac{-15\cdot3^{\frac{2}{3}}-18\cdot3^{\frac{1}{3}}+42}{-159}\\ & =\frac{5\cdot3^{\frac{2}{3}}+6\cdot3^{\frac{1}{3}}-14}{53} \end{align*}(0)-2
\[ a^{3}+b^{3}+c^{3}-3abc=\left(a+b+c\right)\left(a^{2}+b^{2}+c^{2}-ab-bc-ca\right) \] なので、\begin{align*} \frac{1}{2\cdot3^{\frac{2}{3}}+3\cdot3^{\frac{1}{3}}+2} & =\frac{\left(\left(2\cdot3^{\frac{2}{3}}\right)^{2}+\left(3\cdot3^{\frac{1}{3}}\right)^{2}+2^{2}-2\cdot3^{\frac{2}{3}}\cdot3\cdot3^{\frac{1}{3}}-3\cdot3^{\frac{1}{3}}\cdot2-2\cdot2\cdot3^{\frac{2}{3}}\right)}{\left(2\cdot3^{\frac{2}{3}}+3\cdot3^{\frac{1}{3}}+2\right)\left(\left(2\cdot3^{\frac{2}{3}}\right)^{2}+\left(3\cdot3^{\frac{1}{3}}\right)^{2}+2^{2}-2\cdot3^{\frac{2}{3}}\cdot3\cdot3^{\frac{1}{3}}-3\cdot3^{\frac{1}{3}}\cdot2-2\cdot3^{\frac{2}{3}}-2\cdot3\cdot3^{\frac{1}{3}}\right)}\\ & =\frac{12\cdot3^{\frac{1}{3}}+9\cdot3^{\frac{2}{3}}+4-18-6\cdot3^{\frac{1}{3}}-4\cdot3^{\frac{2}{3}}}{\left(\left(2\cdot3^{\frac{2}{3}}\right)^{3}+\left(3\cdot3^{\frac{1}{3}}\right)^{3}+2^{3}\right)-3\cdot2\cdot3^{\frac{2}{3}}\cdot3\cdot3^{\frac{1}{3}}\cdot2}\\ & =\frac{5\cdot3^{\frac{2}{3}}+6\cdot3^{\frac{1}{3}}-14}{\left(72+81+8\right)-108}\\ & =\frac{5\cdot3^{\frac{2}{3}}+6\cdot3^{\frac{1}{3}}-14}{53} \end{align*}
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\]
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\frac{1}{2\sqrt{2}+\sqrt{3}+1}
\]
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