Question

Tính hợp lí nếu có thể

b) \(\dfrac{6^3+3.6^2+3^3}{-13}\)

a)  \(\dfrac{20^5.5^{10}}{100^5}\)

Answer 1

b:\(=\dfrac{3^3\cdot2^3+3\cdot2^2\cdot3^2+3^3}{-13}=\dfrac{3^3\left(2^3+2^2+1\right)}{-13}=-3^3=-27\)

a: \(=\dfrac{5^5\cdot4^5\cdot5^{10}}{10^{20}}=\dfrac{5^{15}\cdot2^{10}}{5^{20}\cdot2^{20}}=\dfrac{1}{5^5\cdot2^{10}}\)

Answer 2

b) \(\dfrac{6^3+3.6^2+3^3}{-13}=\dfrac{2^3.3^3+3.3^2.2^2+3^3}{-13}=\dfrac{3^3\left(2^3+2^2+1\right)}{-13}\)

\(=\dfrac{3^3.\left(8+4+1\right)}{-13}=\dfrac{3^3.13}{-13}=\dfrac{27}{-1}=-27\)

a)\(\dfrac{20^5.5^{10}}{100^5}=\dfrac{20^5.5^{10}}{20^5.5^5}=\dfrac{5^5}{1}=3125\)