Question

Tính :

a) 25\(^{10}\). \(\left(\dfrac{1}{5}\right)^{20}\)+ \(\left(-\dfrac{3}{4}\right)^8\). \(\left(-\dfrac{4}{3}\right)^8\)- 2011\(^0\)

Answer 1

\(25^{10}\cdot\dfrac{1}{5^{20}}+\left(-\dfrac{3}{4}\right)^8\cdot\left(-\dfrac{4}{3}\right)^8-2011^0\\ =\left(5^2\right)^{10}\cdot\dfrac{1}{5^{20}}+\dfrac{\left(-3\right)^8}{4^8}\cdot\dfrac{\left(-4\right)^8}{3^8}-1\\ =5^{20}\cdot\dfrac{1}{5^{20}}+\dfrac{\left(-3\right)^8}{4^8}\cdot\dfrac{\left(-4\right)^8}{3^8}-1\\ =\dfrac{5^{20}}{5^{20}}+\dfrac{\left(-3\right)^8\cdot\left(-4^8\right)}{4^8\cdot3^8}-1\\ =1+1-1\\ =1\)