Question

Bài 1 :

a) \(\dfrac{4^5.4^2}{16^4}\)

b) \(\dfrac{2^8.9^4}{6^6.8^3}\)

c) \(\dfrac{6^3+3.6^2+3^3}{-13}\)

d) \(E=\dfrac{1}{3}+\left(\dfrac{+1}{3}\right)^2+\left(\dfrac{1}{3}\right)^3+...+\left(\dfrac{1}{3}\right)^{100}\)

e) \(\dfrac{\left(0,9\right)^7}{\left(0,3\right)^8}\)

Answer 1

A) \(\dfrac{4^5.4^2}{16^4}=\dfrac{4^7}{\left(2^4\right)^4}=\dfrac{2^{14}}{2^{16}}=\dfrac{1}{4}\)

b)\(\dfrac{2^8.9^4}{6^6.8^3}=\dfrac{2^8.\left(3^2\right)^4}{2^6.3^6.\left(2^3\right)^3}=\dfrac{2^8.3^8}{2^{15}.3^6}=\dfrac{9}{128}\)

Answer 2

c) \(\dfrac{6^3+3.6^2+3^3}{-13}=\dfrac{2^3.3^3+3.2^2.3^2+3^3}{-13}=\dfrac{2^3.3^3+3^3.2^2+3^3}{-13}=\dfrac{3^3.\left(2^3+2^2+1\right)}{-13}=\dfrac{3^3.13}{-13}=-9\)

Answer 3

Tự làm tiếp đi nhé , 2 câu sau dạng cơ bản , hồi lớp 6 học rồi

Answer 4

\(\text{a) }\dfrac{4^5\cdot4^2}{16^4}=\dfrac{4^7}{\left(4^2\right)^4}=\dfrac{4^7}{4^8}=\dfrac{1}{4}\)

\(\text{b) }\dfrac{2^8\cdot9^4}{6^6\cdot8^3}\\ =\dfrac{2^8\cdot\left(3^2\right)^4}{\left(2\cdot3\right)^6\cdot\left(2^3\right)^3}\\ =\dfrac{2^8\cdot3^8}{2^6\cdot3^6\cdot2^9}\\ =\dfrac{2^8\cdot3^8}{2^{15}\cdot3^6}\\ =\dfrac{3^2}{2^7}\\ =\dfrac{9}{128}\)

\(\text{c) }\dfrac{6^3+3\cdot6^2+3^3}{-13}\\ =\dfrac{6^3+3\cdot6^2+3^3}{-13}\\ =\\ \dfrac{\left(2\cdot3\right)^3+3\cdot\left(2\cdot3\right)^2+3^3}{-13}\\ =\dfrac{2^3\cdot3^3+3\cdot2^2\cdot3^2+3^3}{-13}\\ =\dfrac{8\cdot3^3+4\cdot3^3+3^3}{-13}\\ =\dfrac{3^3\cdot\left(8+4+1\right)}{-13}\\ =\dfrac{3^3\cdot13}{-13}\\ =-3^3\\ =-27\)

\(\text{d) }E=\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{3}\right)^3+...+\left(\dfrac{1}{3}\right)^{100}\\ 3E=3\left[\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{3}\right)^3+...+\left(\dfrac{1}{3}\right)^{100}\right]\\ 3E=1+\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2+...+\left(\dfrac{1}{3}\right)^{99}\\ 3E-E=\left[1+\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2+...+\left(\dfrac{1}{3}\right)^{99}\right]-\left[\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{3}\right)^3+...+\left(\dfrac{1}{3}\right)^{100}\right]\\ 2E=1-\left(\dfrac{1}{3}\right)^{100}\\ 2E=1-\dfrac{1}{3^{100}}\\ E=\dfrac{1-\dfrac{1}{3^{100}}}{2}\)

\(\text{e) }\dfrac{\left(0.9\right)^7}{\left(0.3\right)^8}\\ =\dfrac{\left(0.3\cdot3\right)^7}{\left(0.3\right)^8}\\ =\dfrac{\left(0.3\right)^7\cdot3^7}{\left(0.3\right)^8}\\ =\dfrac{3^7}{0.3}=\dfrac{3^6}{0.1}\\ =\dfrac{729}{\dfrac{1}{10}}\\ =729\cdot10\\ =7290\)