Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Mk lm tắt nha
Hk bik có đúng hk nữa
A= 1/2-1/12024
.... Tự tính kết quả nha
gọi A = 1/2+1/4+1/8+......+1/1024 2 x A = 1 + 1/2 + 1/4+......+1/1024 2 x A - A = (1+1/2+1/4 +....+1/512) - (1/2+1/4+1/8+.....1/1024) A= 1-1/1024 = 1023/1024 vậy A= 1023/1024
A x 2 = 1 - ( 1/2 + 1/4 + 1/8 + 1/16 + ..... + 1/512 + 1/1024 ) - 1/1024
A x 2 = 1 - 1/1024 + A
A x 2 - A = 1 - 1/1024
A = 1 - 1/1024
A = 1023 /1024
Đặt tổng trên là A. Ta có
A x 2 = 1+ 1/2+1/4+1/8+ 1/16+1/32+ 1/64+ 1/128 + 1/256 + 1/512
Ax2 - A = 1+ 1/2+1/4+1/8 +1/16 + 1/32 +1/64+ 1/128 + 1/256+ 1/512 - ( 1/2 + 1/4 +1/8+1/16+1/32+1/64 + 1/128+ 1/256 + 1/512+ 1/1024)
A = 1+ 1/2 +1/4+1/8+1/16+1/32+1/64+1/128+1/256 + 1/512 - 1/2-1/4-1/8-1/16-1/32-1/64-1/128-1/256-1/512- 1/1024
A = 1 - 1/ 1024 = 1023/1024
A = 1 - 1/2 + 1/2 - 1/4 + 1/4 - 1/8 + ..... + 1/256 - 1/512 + 1/512 - 1/1024
A = 1 - 1/1024
A = 1023/1024
Ta có: A=1/2 + 1/4 + 1/8 +...+ 1/512 + 1/1024 2A= 1+ 1/2 + 1/4+...+1/256+1/512 2A - A=(1+1/2+1/4+...+1/256+1/512) - ( 1/2+1/4+1/8+...+1/512+1/1024) suy ra A= 1 - 1/1024= 1023/1024 . Vậy A = 1023/1024. nhớ chọn câu trả lời của mình nha!
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{512}+\frac{1}{1024}\)
Đặt tổng trên là A , ta có :
\(A\times2=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\right)\)
\(A\times2=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\right)\times2\)
\(A\times2=\frac{1}{2}\times2+\frac{1}{4}\times2+\frac{1}{8}\times2+...+\frac{1}{512}\times2+\frac{1}{1024}\times2\)
\(A\times2=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)
\(A\times2-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\right)\)
\(A=\left(1-\frac{1}{1024}\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{1}{8}-\frac{1}{8}\right)+....+\left(\frac{1}{256}-\frac{1}{256}\right)+\left(\frac{1}{1024}-\frac{1}{1024}\right)\)
\(A=\frac{1023}{1024}\)
Nguyễn Linh Chi : Mong cô check bài này cho em ạ !!
\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{512}+\dfrac{1}{1024}\)
\(=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)
\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)
\(\Rightarrow2A-A=A=1-\dfrac{1}{2^{10}}\)
=1-(1/2+1/2-1/4+1/4-1/8+1/8...-1/1024+1/1024-1/1024)
=1-1/1024
=1023/1024
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)
\(=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)
Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}+\frac{1}{2^9}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
\(\Rightarrow A=1-\frac{1}{2^{10}}=\frac{2^{10}-1}{2^{10}}\)
vậy \(A=\frac{2^{10}-1}{2^{10}}\)
đề phải là 1 +1/2 + 1/4 +1/32 + 1/64 + 1/128 +1/256 +/512 +1/1024 moi dug