Violympic toán 7

GJ

1.Tính :

a) 1253 : 57

b) \(\left(\dfrac{2}{7}\right)^{18}:\left(\dfrac{4}{49}\right)^5:\left(\dfrac{8}{343}\right)^2\)

c) 3 - \(\left(-\dfrac{7}{9}\right)^0+\left(\dfrac{1}{3}\right)^5.3^5\)

d) \(\dfrac{45^{10}.5^{20}}{75^{15}}\)

H24
10 tháng 10 2018 lúc 16:28

a, \(125^3:5^7=\left(5^3\right)^3:5^7=5^9:5^7=5^2\)

b, \(\left(\dfrac{2}{7}\right)^{18}:\left(\dfrac{4}{49}\right)^5:\left(\dfrac{8}{343}\right)^2\)

= \(\left(\dfrac{2}{7}\right)^{18}:\left(\dfrac{2^2}{7^2}\right)^5:\left(\dfrac{2^3}{7^3}\right)^2\)

= \(\left(\dfrac{2}{7}\right)^{18}:\left[\left(\dfrac{2}{7}\right)^2\right]^5:\left[\left(\dfrac{2}{7}\right)^3\right]^2\)

=\(\left(\dfrac{2}{7}\right)^{18}:\left(\dfrac{2}{7}\right)^{10}:\left(\dfrac{2}{7}\right)^6\)

= \(\left(\dfrac{2}{7}\right)^{18-10-6}=\left(\dfrac{2}{7}\right)^2\)

c, \(3-\left(\dfrac{-7}{9}\right)^0+\left(\dfrac{1}{3}\right)^5.3^5\)

= 3 - 1 +\(\left[\left(\dfrac{1}{3}\right)^5.3^5\right]\)

= 2 + 1=3

d, \(\dfrac{45^{10}.5^{20}}{75^{15}}=\dfrac{\left(9.5\right)^{10}.5^{20}}{\left(25.3\right)^{15}}=\dfrac{\left(3^2\right)^{10}.5^{10}.5^{20}}{\left(5^2\right)^{15}.3^{15}}\)

= \(\dfrac{3^{20}.5^{30}}{5^{30}.3^{15}}=3^5\)

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