K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

11 tháng 8 2018

\(\left(5-\frac{43}{10}\right)-\left(\frac{42}{19}-\left(\frac{7}{2}-\frac{59}{10}\right)\right)+\frac{42}{19}+\frac{59}{10}+\frac{4}{5}\)

\(=5-\frac{43}{10}-\left(\frac{42}{19}-\frac{7}{2}+\frac{59}{10}\right)+\frac{42}{19}+\frac{59}{10}+\frac{4}{5}\)

\(=5-\frac{43}{10}-\frac{42}{19}+\frac{7}{2}-\frac{59}{10}+\frac{42}{19}+\frac{59}{10}+\frac{4}{5}\)

\(=5-\frac{43}{10}+\frac{7}{2}+\frac{4}{5}\)

\(=5+\frac{35}{10}+\frac{8}{10}-\frac{43}{10}=5\)

27 tháng 6 2016

làm giúp mk đi mà chiều mk phải nộp rồi

3 tháng 10 2017

\(10^6-5^7\)

= \(\left(2.5\right)^6-5^7\)

= \(2^6.5^6-5^7\)

= \(5^6.\left(2^6-5\right)\)

= \(5^6.59\) \(⋮\) \(59\)

\(\Rightarrow\left(10^6-5^7\right)⋮59\)

7 tháng 10 2017

Ta có 10\(^6\) - 5\(^7\) = 2\(^6\) . 5\(^6\) - 5\(^7\)

= 5\(^6\) . (2\(^6\) - 5)

= 5\(^6\) . (6\(^4\) - 5)

= 5\(^6\) . 5\(^9\) chia hết cho 59

Vậy 10\(^6\) - 5\(^7\) chia hết cho 59

5 tháng 1 2019

có giải đấy

người nhanh nhất và chính xác 

thưởng 3

5 tháng 1 2019

\(\frac{-3}{7}+\frac{5}{9}:\frac{-35}{18}\cdot\left(\frac{-34}{59}\right)^0+\left(-1\right)^{2017}\)

\(=\frac{-3}{7}+\frac{5}{9}\cdot\frac{-18}{35}\cdot1+\left(-1\right)\)

\(=\frac{-3}{7}+\frac{-2}{7}+\left(-1\right)\)

\(=\frac{-5}{7}+\left(-1\right)\)

\(=\frac{-5}{7}+\frac{-7}{7}=\frac{-12}{7}\)

29 tháng 12 2021

a: \(=\dfrac{-5}{8}-2\cdot\dfrac{5}{2}+\dfrac{1}{2}=-\dfrac{5}{8}-5+\dfrac{1}{2}=-\dfrac{41}{8}\)

23 tháng 6 2017

1.

\(\frac{3}{10}-\left[\left(-\frac{5}{6}\right)-\left(-\frac{2}{3}\right)\right]\)

\(=\frac{3}{10}-\left(-\frac{13}{30}\right)=\frac{11}{15}\)

2.

\(\frac{9}{10}+\left(-\frac{7}{8}\right)-\left(-\frac{2}{5}\right)-\frac{4}{3}\)

\(=\frac{9}{10}-\frac{7}{8}+\frac{2}{5}-\frac{4}{3}=-\frac{109}{120}\)

1 tháng 7 2018

\(A=\left(\frac{5}{3}-\frac{3}{7}+9\right)-\left(2+\frac{5}{7}-\frac{2}{3}\right)+\left(\frac{8}{7}-\frac{4}{3}-10\right)\)

\(A=\frac{5}{3}-\frac{3}{7}+9-2-\frac{5}{7}+\frac{2}{3}+\frac{8}{7}-\frac{4}{3}-10\)

\(A=\left(\frac{5}{3}+\frac{2}{3}-\frac{4}{3}\right)-\left(\frac{3}{7}+\frac{5}{7}-\frac{8}{7}\right)+\left(9-2-10\right)\)

\(A=1-1-3\)

\(A=-3\)

Vậy \(A=-3\)

1 tháng 7 2018

\(A=\left(\frac{3}{5}-\frac{3}{7}+9\right)-\left(2+\frac{5}{7}-\frac{2}{3}\right)+\left(\frac{8}{7}-\frac{4}{3}-10\right)\)

\(A=\frac{5}{3}-\frac{3}{7}+9-2-\frac{5}{7}+\frac{2}{3}+\frac{8}{7}-\frac{4}{3}-10\)

\(A=\left(\frac{5}{3}+\frac{2}{3}-\frac{4}{3}\right)-\left(\frac{3}{7}+\frac{5}{7}-\frac{8}{7}\right)+\left(9-2-10\right)\)

\(A=1+0-3\)

\(A=-2\)

Vậy \(A=-2\)

17 tháng 9 2020

a) \(4\frac{5}{9}:\left(-\frac{5}{7}\right)+\frac{49}{9}:\left(-\frac{5}{7}\right)=\frac{41}{9}:\left(-\frac{5}{7}\right)+\frac{49}{9}:\left(-\frac{5}{7}\right)\)

\(=\frac{41}{9}\cdot\left(-\frac{7}{5}\right)+\frac{49}{9}\cdot\left(-\frac{7}{5}\right)=\left(\frac{41}{9}+\frac{49}{9}\right)\cdot\left(-\frac{7}{5}\right)=10\cdot\left(-\frac{7}{5}\right)=-14\)

b) \(\left(\frac{-3}{5}+\frac{4}{9}\right):\frac{7}{11}+\left(\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)

\(=\left(\frac{-3}{5}+\frac{4}{9}+\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)

\(=\left(\frac{-3}{5}+\frac{-2}{5}+\frac{4}{9}+\frac{5}{9}\right):\frac{7}{11}\)

\(=\left(-1+1\right):\frac{7}{11}=0\cdot\frac{11}{7}=0\)

c) \(\left(\frac{3}{4}\right)^4\cdot\left(\frac{8}{9}\right)^2=\left(\frac{3}{4}\right)^2\cdot\left(\frac{3}{4}\right)^2\cdot\left(\frac{8}{9}\right)^2=\left(\frac{3}{4}\cdot\frac{3}{4}\cdot\frac{8}{9}\right)^2\)

\(=\left(\frac{1}{2}\right)^2=\frac{1}{4}\)

d) \(\left(-\frac{3}{5}\right)^6\cdot\left(-\frac{5}{3}\right)^5=\left(-\frac{3}{5}\right)^5\cdot\left(-\frac{3}{5}\right)\cdot\left(-\frac{5}{3}\right)^5=\left[\left(-\frac{3}{5}\right)\cdot\left(-\frac{5}{3}\right)\right]^5\cdot\left(-\frac{3}{5}\right)\)

\(=1^5\cdot\left(-\frac{3}{5}\right)=1\cdot\left(-\frac{3}{5}\right)=-\frac{3}{5}\)

e) \(\frac{8^{14}}{4^4\cdot64^5}=\frac{\left(2^3\right)^{14}}{\left(2^2\right)^4\cdot\left(2^6\right)^5}=\frac{2^{42}}{2^8\cdot2^{30}}=\frac{2^{42}}{2^{38}}=2^4=16\)

f) \(\frac{9^{10}\cdot27^7}{81^7\cdot3^{15}}=\frac{\left(3^2\right)^{10}\cdot\left(3^3\right)^7}{\left(3^4\right)^7\cdot3^{15}}=\frac{3^{20}\cdot3^{21}}{3^{28}\cdot3^{15}}=\frac{3^{41}}{3^{43}}=3^{-2}=\frac{1}{3^2}=\frac{1}{9}\)