Найдите значение выражения: 1/27а^-1 b^-5 * 81a^2 b^4 если а= 1/7 b=1/14. Created by sidelnikovan1. algebra-ru

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$\displaystyle \frac{1}{27}a^{-1}b^{-5}*81a^{2}b^{4} = \frac{81a^{2}b^{4}}{27ab^{5}} = \frac{3a}{b} \\ \\ npu:a= \frac{1}{7};b= \frac{1}{14} \\ \\ \frac{3a}{b}= \frac{3}{7}: \frac{1}{14}= \frac{3}{7}*14=6$

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$\frac{1}{27} a^{-1 }b^{-5} * 81a^2 b^4 = \frac{81}{27}*a^{-1+2}*b^{-5+4}=3ab^{-1}$
если $a= \frac{1}{7} ,$ $b= \frac{1}{14}$

$3ab^{-1}=3* \frac{1}{7} *( \frac{1}{14} )^{-1}= \frac{3}{7} *14=3*2=6$

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