PT
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
23 tháng 9 2021
Sửa đề :
\(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
Bài làm :
\(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(=\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+...+\frac{1}{128}-\frac{1}{256}\)
\(=\frac{1}{4}-\frac{1}{256}=\frac{63}{256}\)
8 tháng 3 2020
= 128/256 + 64/256 + 32/256 + 16/256 + 8/256 + 4/256 + 2/256 + 1/256
= 255/256
TD
6 tháng 11 2019
A= 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256
2A= 2(1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256)
= 1+1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128
=>A = 2A-A =1+1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 -1/2 - 1/4 - 1/8 - 1/16 - 1/32 - 1/64 - 1/128 - 1/256
=1-1/256
=255/256
TL
7 tháng 4 2018
Mk có cách giải khác nè
1/4+1/8+1/16+1/32+1/64+1/128
= 1/2-1/4+1/4-1/8+1/8-1/16+1/16-1/32+1/32-1/64+1/64-1/128
= 1/2-1/128
= 63/128
17 tháng 9 2021
\(\frac{1}{2}\)+ \(\frac{1}{4}\)+ \(\frac{1}{16}\)+ \(\frac{1}{32}\)+ \(\frac{1}{64}\)+ \(\frac{1}{128}\)= \(\frac{123}{234}\)
1 tháng 8 2017
a ) 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256
Đạt A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256
A x 2 = 2 x ( 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256)
A x 2 = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128
Lấy A x 2 - A ta có :
A x 2 - A = 1 + 1/2 + ..... + 1/128 - 1/2 + 1/4 + ........ + 1/256
A x ( 2 - 1 ) = 1 - 1/ 256
A = 255/256
b) 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
Đặt A = 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
A x 3 = 3 x ( 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729)
= 1 + 1/ 3 + 1/9 + 1/27 + 1/81 + 1/243
Lấy A x 3 - A ta có :
A x 3 - A = 1 + 1/3 + 1/9 +..... + 1/243 - 1/3 + 1/9 +........+ 1/243 + 1/29
A x ( 3 - 1 ) = 1 - 1/29
A x2 = 28/29
A = 28/29 : 2 ( tự tính
1 tháng 8 2017
Đặt A = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.....+\frac{1}{256}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{128}\)
\(\Rightarrow2A-A=1-\frac{1}{256}\)
\(\Rightarrow A=\frac{255}{256}\)
TM
4 tháng 7 2016
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(\Rightarrow2A=\frac{2}{2}+\frac{2}{4}+\frac{2}{8}+\frac{2}{16}+\frac{2}{32}+\frac{2}{64}+\frac{2}{128}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\right)\)
\(\Rightarrow A=1-\frac{1}{128}=\frac{128}{128}-\frac{1}{128}=\frac{127}{128}\)
4 tháng 7 2016
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{128}\)
\(=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+.....+\left(\frac{1}{64}-\frac{1}{128}\right)\)
\(=1-\frac{1}{128}=\frac{127}{128}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
28 tháng 5 2023
A = \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\)+ \(\dfrac{1}{32}\)+\(\dfrac{1}{64}\)+\(\dfrac{1}{128}\)
A\(\times\) 2 = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\)+ \(\dfrac{1}{32}\)+ \(\dfrac{1}{64}\)
A \(\times\) 2 - A = 1 - \(\dfrac{1}{128}\)
A\(\times\)(2-1) = \(\dfrac{128-1}{128}\)
A = \(\dfrac{127}{128}\)
BD
28 tháng 5 2023
Gọi \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\) là B
\(B=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\)
\(2\cdot B=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{12}+\dfrac{1}{32}+\dfrac{1}{64}\)
\(2\cdot B-B=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{12}+\dfrac{1}{32}+\dfrac{1}{64}-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\right)\)
\(B=1+\left(\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+.....+\dfrac{1}{64}-\dfrac{1}{64}\right)-\dfrac{1}{128}\)
\(B=1+0-\dfrac{1}{128}\)
\(B=1-\dfrac{1}{128}\)
\(B=\dfrac{128}{128}-\dfrac{1}{128}\)
\(B=\dfrac{127}{128}\)
tính nhanh p/s 1+ 5/4 + 5/8 + 5/16 + 5/32 + 5/64
9 tháng 5 2023
b: A=1/3+1/9+...+1/3^10
=>3A=1+1/3+...+1/3^9
=>A*2=1-1/3^10=(3^10-1)/3^10
=>A=(3^10-1)/(2*3^10)
c: C=3/2+3/8+3/32+3/128+3/512
=>4C=6+3/2+...+3/128
=>3C=6-3/512
=>C=1023/512
d: A=1/2+...+1/256
=>2A=1+1/2+...+1/128
=>A=1-1/256=255/256
Answer:
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=\frac{64}{128}+\frac{32}{128}+\frac{16}{128}+\frac{8}{128}+\frac{4}{128}+\frac{2}{128}+\frac{1}{128}\)
\(=\frac{127}{128}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=\left(\frac{1}{2}+\frac{1}{4}\right)+\left(\frac{1}{8}+\frac{1}{16}\right)+\left(\frac{1}{32}+\frac{1}{64}\right)+\frac{1}{128}\)
\(=\frac{3}{4}+\frac{3}{16}+\frac{3}{64}+\frac{1}{128}\)
\(=\frac{96}{128}+\frac{24}{128}+\frac{6}{128}+\frac{1}{128}\)
\(=\frac{127}{128}\)