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NG
1
1 tháng 3 2015
Ta có:
\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\)
\(\Rightarrow2A=1+\frac{1}{2}+...+\frac{1}{2^9}\)
Lấy \(2A-A\), ta có:
\(2A-A=A=\left(1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
\(=1+\frac{1}{2}+...+\frac{1}{2^9}-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{10}}\)
\(=\left(1-\frac{1}{2^{10}}\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+...+\left(\frac{1}{2^9}-\frac{1}{2^9}\right)\)
\(=1-\frac{1}{2^{10}}\)
\(=1-\frac{1}{1024}\)
\(=\frac{1023}{1024}\)
Vậy \(A=\frac{1023}{1024}\)
17 tháng 2 2022
a: \(=\dfrac{17}{7}+\dfrac{2}{9}-\dfrac{10}{7}-\dfrac{5}{3}\cdot9=1+\dfrac{2}{9}-15=-14+\dfrac{2}{9}=-\dfrac{126}{9}+\dfrac{2}{9}=-\dfrac{124}{9}\)
b: \(=\dfrac{-11}{23}\left(\dfrac{6}{7}+\dfrac{8}{7}\right)-\dfrac{1}{23}=\dfrac{-22}{23}-\dfrac{1}{23}=-1\)
c: \(=\left(\dfrac{377}{-231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot\dfrac{4-3-1}{24}=0\)
d: \(=\dfrac{12}{7}\left(19+\dfrac{5}{8}-15-\dfrac{1}{4}\right)=\dfrac{12}{7}\cdot\dfrac{35}{8}=\dfrac{15}{2}\)
30 tháng 1 2020
Ta có: \(2^{10}=1024\)
Đặt \(A=2+2^2+...+2^{10}\)và \(B=\frac{1023}{2+2^2+...+2^{10}}\)
\(2A=2^2+2^3+...+2^{11}\)
\(A=2^{11}-2\)
Thay A vào B, ta có: \(B=\frac{2^{10}-1}{2^{11}-2}=\frac{2^{10}-1}{2\left(2^{10}-1\right)}=\frac{1}{2}\)
Vậy B= 1/2
28 tháng 2 2018
\(=\frac{12}{7}\cdot\frac{3}{4}-\frac{6}{7}\cdot\frac{4}{3}+\frac{6}{7}\)
\(=\frac{6}{7}\left(\frac{3}{2}-\frac{4}{3}+1\right)\)
\(=\frac{6}{7}\left(\frac{1}{6}+1\right)=\frac{6}{7}\cdot\frac{7}{6}=1\)
2.
\(=2017\cdot2018\cdot\left[\left(2016\cdot2018\right)-\left(2016\cdot2017\right)\right]\)
\(=2017\cdot2018\cdot2016\left(2018-2017\right)=2016\cdot2017\cdot2018\)
3.
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)....\left(\frac{1}{100}-1\right)=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot....\cdot\frac{99}{100}\)
\(=\frac{1}{100}\)
4.
\(=\frac{1+2+2^2+2^4+...+2^9}{2\left(1+2+2^2+2^3+2^4+...+2^9\right)}\)
\(=\frac{1}{2}\)
28 tháng 2 2018
mình chỉ làm được câu 3 thôi
có \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)....\left(\frac{1}{100}-1\right)\)
\(=\frac{-1}{2}\times\frac{-2}{3}\times....\times\frac{-99}{100}\)
\(=\frac{\left(-1\right)\left(-2\right)....\left(-99\right)}{2\times3\times....\times100}\)
\(=\frac{-\left(1\times2\times....\times99\right)}{2\times3\times....\times100}\)
\(=\frac{-1}{100}\)
3 tháng 1 2023
a: =-45*24+120
=-1080+120
=-960
b: =134(-1+51-48)
=134*2
=268
c: =-41*59-41*2+59*41-59*2
=-2(41+59)
=-200
NL
0
21 tháng 3 2020
\(C=\left(\frac{1}{2}-1\right)+\left(1-\frac{3}{4}\right)+\left(\frac{7}{8}-1\right)+...+\left(1-\frac{1023}{1024}\right)\)
\(C=\left(\frac{1}{2^1}-\frac{2}{2}\right)+\left(\frac{2^2}{2^2}-\frac{3}{2^2}\right)+...+\left(\frac{1024}{1024}-\frac{1023}{2^{10}}\right)\)
\(C=\frac{-1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(2C=-1+\frac{1}{2}-\frac{1}{2^2}+...+\frac{1}{2^9}\)
\(2C+C=\left(-1+\frac{1}{2}-\frac{1}{2^2}+...+\frac{1}{9}\right)+\left(-\frac{1}{2}+\frac{1}{2^2}-..+\frac{1}{2^{10}}\right)\)
\(3C=\frac{1}{2^{10}}-1\)
\(C=\frac{\frac{1}{2^{10}}-1}{3}\)
hok tốt!!
11 tháng 5 2023
\(=\dfrac{2}{2}+\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{56}\)
\(=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}\right)\)
=2*7/8=7/4
\(\frac{1023}{2^1+2^2+.........+2^{10}}=\frac{1023}{2\left(1+2+2^2+...........+2^9\right)}\)
\(A=1+2+2^2+...............+2^9\)
\(2A=2+2^2+2^3+...........2^{10}\)
\(2A-A=\left(2+2^2+2^3+..............+2^{10}\right)-\left(1+2+2^2+...........+2^9\right)\)
\(A=2^{10}-1=1023\)
\(\Rightarrow A=2^{10}-1=\frac{1023}{2.1023}=\frac{1}{2}\)