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20 tháng 10 2022

\(=\left(\dfrac{3}{7}\right)^5\cdot\left(\dfrac{3}{7}\right)\cdot\left(\dfrac{5}{3}\right)^6:\left(\dfrac{5^4}{7^3}\right)^2\)

\(=\dfrac{3^6}{7^6}\cdot\dfrac{5^6}{3^6}:\dfrac{5^8}{7^6}\)

\(=\dfrac{5^6}{5^8}=\dfrac{1}{25}\)

\(=\left(\dfrac{3}{7}\right)^5\cdot\left(\dfrac{3}{7}\right)\cdot\left(\dfrac{5}{3}\right)^6:\left(\dfrac{625}{343}\right)^2\)

\(=\left(\dfrac{3}{7}\cdot\dfrac{5}{3}\right)^6:\left(\dfrac{625}{343}\right)^2\)

\(=\dfrac{5^6}{7^6}:\dfrac{5^8}{7^6}=\dfrac{1}{25}\)

a: \(=\dfrac{3^3\cdot2^6}{3^{-4}\cdot2^6}=3^7\)

b: \(=\left(\dfrac{3}{7}\right)^5\cdot\left(\dfrac{3}{7}\right)\cdot\dfrac{5^6}{3^6}:\left(\dfrac{625}{343}\right)^2\)

\(=\dfrac{3^6}{7^6}\cdot\dfrac{5^6}{3^6}:\dfrac{5^8}{7^6}\)

\(=\dfrac{1}{5^2}\)

c: \(=5^{4+3}\cdot\left(\dfrac{5}{2}\right)^{-5}\cdot\dfrac{1}{25}\)

\(=5^5\cdot\left(\dfrac{2}{5}\right)^5=2^5\)

a: \(=\dfrac{3^3\cdot2^6}{3^{-4}\cdot2^6}=3^7\)

b: \(=\left(\dfrac{3}{7}\cdot\dfrac{5}{3}\right)^6\cdot\dfrac{5}{3}\cdot\dfrac{3}{7}:\left(\dfrac{7^3}{5^4}\right)^{-2}\)

\(=\left(\dfrac{5}{7}\right)^6\cdot\dfrac{5}{7}\cdot\left(\dfrac{5}{7}\right)^6\cdot5^2\)

\(=\left(\dfrac{5}{7}\right)^{13}\cdot5^2\)

c: \(=5^4\cdot2.5^{-5}\cdot125\cdot0.04\)

\(=5^4\cdot5\cdot\left(\dfrac{5}{2}\right)^{-5}\)

\(=5^5\cdot\left(\dfrac{2}{5}\right)^5=2^5\)

30 tháng 11 2017

Các bạn trả lời giúp mk nha. Mk đang cần gấp. Chều nay mk kiểm tra rồi

30 tháng 11 2017

0 cần trả lời hết cũng đc

tính a) \(\left[\dfrac{0.8\div\left(\dfrac{4}{5}\cdot1025\right)}{0.64-1}+\dfrac{\left(1.08-\dfrac{2}{25}\right)\div\dfrac{4}{7}}{\left(6\dfrac{5}{7}-3\dfrac{1}{4}\right)\cdot2\dfrac{2}{17}}+\left(1.2\cdot0.5\right)\div\dfrac{4}{5}\right]\) b) \(\left(0.2\right)^{-3}\left[\left(-\dfrac{1}{5}\right)^{-2}\right]^{-1}+\left[\left(\dfrac{1}{2}\right)^{-3}\right]^{-2}\div\left(2^{-3}\right)^{-1}-\left(0.175\right)^{-2}\) c) \(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\) d)...
Đọc tiếp

tính

a) \(\left[\dfrac{0.8\div\left(\dfrac{4}{5}\cdot1025\right)}{0.64-1}+\dfrac{\left(1.08-\dfrac{2}{25}\right)\div\dfrac{4}{7}}{\left(6\dfrac{5}{7}-3\dfrac{1}{4}\right)\cdot2\dfrac{2}{17}}+\left(1.2\cdot0.5\right)\div\dfrac{4}{5}\right]\)

b) \(\left(0.2\right)^{-3}\left[\left(-\dfrac{1}{5}\right)^{-2}\right]^{-1}+\left[\left(\dfrac{1}{2}\right)^{-3}\right]^{-2}\div\left(2^{-3}\right)^{-1}-\left(0.175\right)^{-2}\)

c) \(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)

d) \(\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{3}\)

e) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2\div2\)

f) \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

g) \(\dfrac{1}{-\left(2017\right)\left(-2015\right)}+\dfrac{1}{\left(-2015\right)\left(-2013\right)}+...+\dfrac{1}{\left(-3\right)\cdot\left(-1\right)}\)

h) \(\left(1-\dfrac{1}{1\cdot2}\right)+\left(1-\dfrac{1}{2\cdot3}+...+\left(1-\dfrac{1}{2017\cdot2018}\right)\right)\)

3
7 tháng 10 2017

c)

Ta có :\(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)

\(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{3}{2}}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{\dfrac{8}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{3}{8}}\) \(=2+\dfrac{1}{\dfrac{11}{8}}\) \(=2+\dfrac{8}{11}\) \(=\dfrac{30}{11}\)

7 tháng 10 2017

d) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)

\(=3-1+\left(\dfrac{1}{2}\right)^2:2\)

\(=3-1+\dfrac{1}{4}:2\)

\(=3-1+\dfrac{1}{8}\)

\(=\dfrac{17}{8}\)

8 tháng 7 2018

\(a)\dfrac{4}{5}-\left(\dfrac{-2}{7}\right)-\dfrac{7}{10}\)

\(=\dfrac{56}{70}+\dfrac{20}{70}+\dfrac{-49}{70}\)

\(=\dfrac{27}{70}\)

\(b)\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)

\(=\left(\dfrac{36}{6}+\dfrac{-4}{6}+\dfrac{3}{6}\right)-\left(\dfrac{30}{6}+\dfrac{10}{6}+\dfrac{-9}{6}\right)-\left(\dfrac{18}{6}+\dfrac{-14}{6}+\dfrac{15}{6}\right)\)

\(=\dfrac{35}{6}-\dfrac{31}{6}-\dfrac{19}{6}\)

\(=\dfrac{35+\left(-31\right)+\left(-19\right)}{6}\)

\(=-\dfrac{15}{6}=-2\dfrac{1}{2}\)

9 tháng 7 2018

bài dễ thấy móa lun

hehe

19 tháng 11 2018

5) \(\left(-2\right)^2+\sqrt{36}-\sqrt{9}+\sqrt{25}\)

=\(4+6-3+5\)

=\(12\)

19 tháng 11 2018

2) \(\dfrac{11}{25}.\left(-24,8\right)-\dfrac{11}{25}.75,2\)

=\(\dfrac{11}{25}.\left(-24,8-75,2\right)\)

=\(\dfrac{11}{25}.\left(-100\right)\)

=\(-44\)

10 tháng 10 2018

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

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

11 tháng 2 2018

\(B=\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\frac{2^2}{7^2}-\frac{4}{343}}\)

\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{\frac{8}{2}-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)

\(B=\frac{\frac{343}{343}-\frac{49}{343}+\frac{7}{343}-\frac{1}{343}}{4-\frac{4}{7}+\frac{28}{343}-\frac{4}{343}}\)

\(B=\frac{\frac{300}{343}}{\frac{28}{7}-\frac{4}{7}+\frac{24}{343}}\)

\(B=\frac{\frac{300}{343}}{\frac{24}{7}+\frac{24}{343}}\)

\(B=\frac{\frac{300}{343}}{\frac{1323}{343}+\frac{24}{343}}\)

\(B=\frac{300}{343}:\frac{1347}{343}\)

\(B=\frac{100}{449}\)

11 tháng 2 2018

\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)

\(A=\frac{2^{12}.3^5-2^{12}.3^6}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^6}{5^9.7^3+5^9.2^3.7^3}\)

\(A=\frac{2^{12}.3^5\left(1-3\right)}{2^{12}.3^5.\left(3+1\right)}-\frac{5^{10}.7^3.\left(1-7^3\right)}{5^9.7^3.\left(1+8\right)}\)

\(A=\frac{-2}{4}-\frac{5.\left(-342\right)}{9}\)

\(A=\frac{-1}{2}+\frac{1710}{9}\)

\(A=\frac{-1}{2}+190\)

\(A=\frac{-1}{2}+\frac{380}{2}\)

\(A=\frac{379}{2}\)