K
Khách

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

4 tháng 10 2021

\(\dfrac{4^5.21+4^5.19}{2^{10}.11+2^{10}.5}=\dfrac{4^5.\left(21+19\right)}{2^{10}.\left(11+5\right)}=\dfrac{4^5.30}{2^{10}.16}\)\(=\dfrac{\left(2.2\right)^5.15}{2^{10}.8}=\dfrac{2^5.2^5.15}{2^{10}.8}=\dfrac{15}{8}\)

a: \(\left(0.5\right)^3\cdot2^3=1\)

b: \(\left(0.25\right)^2\cdot16=1\)

c: \(\left(\dfrac{3}{5}\right)^3:\left(-\dfrac{27}{1000}\right)=\dfrac{3^3}{5^3}\cdot\dfrac{-1000}{27}=\dfrac{-1000}{125}=-8\)

\(\Rightarrow A=4.\left[\frac{6}{2.\left(2.4\right)}+\frac{5}{\left(2.4\right).13}+\frac{3}{13.\left(4.4\right)}+\frac{2}{\left(4.4\right).18}+\frac{10}{18.\left(7.4\right)}\right]\)

\(=4.\left(\frac{6}{2.8}+\frac{5}{8.13}+\frac{3}{13.16}+\frac{2}{16.18}+\frac{10}{18.28}\right)=4.\left(\frac{1}{2}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{18}+\frac{1}{18}-\frac{1}{28}\right)\)

\(=4.\left(\frac{1}{2}-\frac{1}{28}\right)=4.\frac{13}{28}=\frac{13}{7}\)

4 tháng 8 2020

Thank you <3

4 tháng 10 2021

\(=\dfrac{2^4\cdot5^4+2^5\cdot5^3}{2^{10}\cdot16}=\dfrac{2^4\cdot5^3\left(5+2\right)}{2^{10}\cdot2^4}=\dfrac{2^4\cdot5^3\cdot7}{2^{14}}=\dfrac{5^3\cdot7}{2^{10}}=\dfrac{875}{1024}\)

1 tháng 4 2019

A=1+(2-3-3+5)+(6-7-8+9)+....+(98-99-100+101)+102

=1+0+0+....+102=103

b) |1-2x|>7

=> 1-2x>7 hoặc 1-2x<-7

=> 2x<-6 hoặc 2x>8

=> x<-3 hoặc x>4

7 tháng 4 2019

\(M=\frac{8^{10}+4^{10}}{8^4+4^{11}}\)

\(M=\frac{8^4+4^{10}.8^6}{8^4+4^{10}.4}\)

\(M=1\frac{8^6}{8^4+4^{10}.4}\)

7 tháng 4 2019

\(M=\frac{8^{10}+4^{10}}{8^4+4^{11}}\)

\(M=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}\)

\(M=\frac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(2^{10}+1\right)}\)

\(M=2^8\)

\(M=256\)

1 tháng 9 2019

\(A=\left(3+\frac{1}{2}-\frac{2}{3}\right)-\left(2-\frac{2}{3}+\frac{5}{2}\right)+\left(-5+\frac{5}{2}-\frac{4}{3}\right)\)

\(=3+\frac{1}{2}-\frac{2}{3}-2+\frac{2}{3}-\frac{5}{2}-5+\frac{5}{2}-\frac{4}{3}\)

\(=\left(3-2-5\right)+\left(\frac{1}{2}-\frac{5}{2}+\frac{5}{2}\right)-\left(\frac{2}{3}-\frac{2}{3}+\frac{4}{3}\right)\)

\(=-4-\frac{1}{2}\)

\(=-\frac{9}{2}\)

\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{72}+\frac{1}{90}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}\)

\(=\frac{9}{10}\)

1 tháng 9 2019

\(A=\left(3+\frac{1}{2}-\frac{2}{3}\right)-\left(2-\frac{2}{3}+\frac{5}{2}\right)+\left(-5+\frac{5}{2}-\frac{4}{3}\right)\)

\(A=3+\frac{1}{2}-\frac{2}{3}-2+\frac{2}{3}-\frac{5}{2}-5+\frac{5}{2}-\frac{4}{3}\)

\(A=\left(3-2-5\right)+\left(\frac{1}{2}-\frac{5}{2}+\frac{5}{2}\right)-\left(\frac{2}{3}-\frac{2}{3}+\frac{4}{3}\right)\)

\(A=-4+\frac{1}{2}-\frac{4}{3}\)

\(A=-\frac{29}{6}\)

1 tháng 3 2019

\(S=1+5+5^2+5^4+...+5^{200}\)

\(\Leftrightarrow5^2S=5^2+5^4+...+5^{202}\)

\(\Leftrightarrow25S=5^2+5^4+...+5^{202}\)

\(\Leftrightarrow25S-S=5^{202}-1\)

\(\Leftrightarrow S=\left(5^{202}-1\right)\div24\)

1 tháng 3 2019

a) S = 1 + 52 + 54 + ... + 5200

=> 52S = 52.(1 + 52 + 54 + ... + 5200)

=> 25S = 52 + 54 + 56 + ... + 5202

=> 25S - S = (52 + 54 + 56 + ... + 5202) - (1 + 52 + 54 + ... + 5200)

=> 24S = 5202 - 1

=> S = \(\frac{5^{202}-1}{24}\)

2 tháng 5 2017

\(\dfrac{\text{45^{10^{ }}}.5^{10}}{75^{10}}=\dfrac{9^{10}.5^{10}.5^{10}}{5^{10}.5^{10}.3^{10}}=\dfrac{9^{10}}{3^{10}}=3^{10}\)

2 tháng 5 2017

\(\dfrac{\left(0,8\right)^5}{\left(0,4\right)^6}=\dfrac{2^5.\left(0,4\right)^5}{\left(0,4\right)^6}=\dfrac{2^5}{0,4}=\dfrac{32}{0,4}=80\)