证明:\( \frac{2^{\frac{1}{2}} \times 3^{\frac{1}{3}} \times 4^{\frac{1}{4}}}{10^{\frac{-1}{5}} \times 5^{\frac{3}{5}}} \div \frac{3^{\frac{4}{3}} \times 5^{\frac{-7}{5}}}{4^{\frac{-3}{5}} \times 6}=10 \)
已知:
\( \frac{2^{\frac{1}{2}} \times 3^{\frac{1}{3}} \times 4^{\frac{1}{4}}}{10^{\frac{-1}{5}} \times 5^{\frac{3}{5}}} \div \frac{3^{\frac{4}{3}} \times 5^{\frac{-7}{5}}}{4^{\frac{-3}{5}} \times 6}=10 \)
要求:
我们需要证明 \( \frac{2^{\frac{1}{2}} \times 3^{\frac{1}{3}} \times 4^{\frac{1}{4}}}{10^{\frac{-1}{5}} \times 5^{\frac{3}{5}}} \div \frac{3^{\frac{4}{3}} \times 5^{\frac{-7}{5}}}{4^{\frac{-3}{5}} \times 6}=10 \)。
解答
我们知道,
$(a^{m})^{n}=a^{m n}$
$a^{m} \times a^{n}=a^{m+n}$
$a^{m} \div a^{n}=a^{m-n}$
$a^{0}=1$
因此,
左边 $=\frac{2^{\frac{1}{2}} \times 3^{\frac{1}{3}} \times 4^{\frac{1}{4}}}{10^{\frac{-1}{5}} \times 5^{\frac{3}{5}}} \div \frac{3^{\frac{4}{3}} \times 5^{\frac{-7}{5}}}{4^{\frac{-3}{5}} \times 6}$
$=\frac{2^{\frac{1}{2}} \times 3^{\frac{1}{3}} \times(2^{2})^{\frac{1}{4}}}{(2 \times 5)^{\frac{-1}{5}} \times 5^{\frac{3}{5}}} \div \frac{3^{\frac{4}{3}} \times 5^{\frac{-7}{5}}}{(2^{2})^{\frac{-3}{5}} \times 2^{1} \times 3^{1}}$
$=\frac{2^{\frac{1}{2}} \times 3^{\frac{1}{3}} \times 2^{2 \times \frac{1}{4}}}{2^{\frac{-1}{5}} \times 5^{\frac{-1}{5}} \times 5^{\frac{3}{5}}} \div \frac{3^{\frac{4}{3}} \times 5^{\frac{-7}{5}}}{2^{\frac{-6}{5}} \times 2^{1} \times 3^{1}}$
$=\frac{2^{\frac{1}{2}} \times 3^{\frac{1}{3}} \times 2^{\frac{1}{2}}}{2^{\frac{-1}{5}} \times 5^{\frac{-1}{5}} \times 5^{\frac{3}{5}}} \times \frac{2^{\frac{-6}{5}} \times 2^{1} \times 3^{1}}{3^{\frac{4}{3}} \times 5^{\frac{-7}{5}}}$
$=2^{\frac{1}{2}+\frac{1}{2}+\frac{1}{5}+\frac{-6}{5}+1} \times 3^{\frac{1}{3}+1-\frac{4}{3}} \times 5^{\frac{1}{5}-\frac{3}{5}+\frac{7}{5}}$
$=2 \frac{5+5-12+10+2}{10} \times 3^{\frac{1+3-4}{3}} \times 5^{\frac{1-3+7}{5}}$
$=2^{\frac{22-12}{10}} \times 3^{\frac{4-4}{3}} \times 5^{\frac{8-3}{5}}$
$=2^{\frac{10}{10}} \times 3^{0} \times 5^{\frac{5}{5}}$
$=2^{1} \times 3^{0} \times 5^{1}$
$=2 \times 1 \times 5$
$=10$
$=$ 右边
证毕。