Question

Вычислить 9^3*27^4/81^5

Answer 1

$\frac{9 ^{3}*27 ^{4} }{81 ^{5} }= \frac{ (3^{2}) ^{3}*( 3^{3}) ^{4} }{( 3^{4}) ^{5} }= \frac{3 ^{6}* 3^{12} }{ 3^{20} }= \frac{3 x^{6+12} }{3 ^{20} }= 3^{18-20}= 3^{-2}= \frac{1}{3 ^{2} }= \frac{1}{9}$

Answer 2

$\frac{9^3*27^4}{81^5} = \frac{(3^2)^3*(3^3)^4}{(3^4)^5} = \frac{ 3^{2*3} * 3^{3*4} }{ 3^{4*5} } = \frac{3^6* 3^{12} }{ 3^{20} } = 3^{6+12-20} = 3^{-2} = \frac{1}{3^2} = \frac{1}{9}$