2047/2048 nha k mình đi mình trả lời đầu tiên đó

1-1/2048=2047/2048

\(\text{Đặt }S=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2048}.\)

\(\Rightarrow2S=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\)

\(\Rightarrow2S-S=S=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2048}\right)\)

\(\Rightarrow S=1-\frac{1}{2048}=\frac{2047}{2048}\)

gọi biểu thức là A

A=1/2+1/4+1/8+...+1/2048=1/2+1/2^2+1/2^3+...+1/2^10

=>2A=1+1/2+1/2^2+...+1/2^9

=>A=2A-A(bạn đặt cột dọc ra rồi sẽ thấy:1/2-1/2=0;1/2^2-1/2^2=0;...)Ta được kết quả bằng 1+1/2^10

Đặt A =1/2 + 1/4 + 1/8 + ...+ 1/1024 + 1/2048

A= 1/2 + 1/2^2 + 1/2^3+...+ 1/2^10 + 1/2^11

2A= 1 +1/2 + 1/2^2 +...+ 1/2^9 + 1/2^10

2A-A= (1 +1/2 + 1/2^2 +...+ 1/2^9 + 1/2^10) - (1/2 + 1/2^2 + 1/2^3+...+ 1/2^10 + 1/2^11)

A= 1+1/2 + 1/2^2 +...+ 1/2^9 + 1/2^10 - 1/2 - 1/2^2 - 1/2^3 - ...- 1/2^10 - 1/2^11

A= 1- 1/2^11

A= 2047/ 2048

Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+...+\frac{1}{1024}+\frac{1}{2048}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+...+\frac{1}{512}+\frac{1}{1024}\)
\(2A-A=1-\frac{1}{2048}\)
\(A=\frac{2047}{2048}\)

Đặt A = 2 + 4 + 8 +  ... + 2048

         = 2 + 2² + 2³ + ... + 2¹¹

=> 2A = 2² + 2³ + 2⁴ + ... + 2¹²

Lấy 2A trừ A theo vế ta có

2A - A = (2² + 2³ + 2⁴ + ... + 2¹²) - (2 + 2² + 2³ + ... + 2¹¹) 

=>   A = 2¹² - 2

Đặt \(A=2+4+8+16+...+1024+2048\)

\(\Rightarrow2A=4+8+16+32+...+2048+4096\)

\(\Rightarrow2A-A=4096-2\)

\(\Rightarrow A=4094\)

các bạn ơi có thể làm theo cách lớp 5 được ko 

vì mình năm nay mới lên lớp 6

Đặt tổng trên là A . Ta có:

\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)

\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)

\(A=1-\frac{1}{1024}\)

\(A=\frac{1023}{1024}\)

Đặt 

\(S=1+2+4+...+2048+4096\)

\(S=1+2^1+2^2+...+2^{11}+2^{12}\)

\(2S=2+2^2+2^3+...+2^{12}+2^{13}\)

\(2S-S=\left(2+2^2+2^3+...+2^{13}\right)-\left(1+2+2^2+..+2^{12}\right)\)

\(S=2^{13}-1=8192-1=8191\)

Gọi A=1+2+4+8+16+...+1024+2048+4096

2A=2+4+8+16+32+...+2048+4096+8192

2A-A=(2+4+8+16+32+...+2048+4096+8192)-(1+2+4+8+16+...+1024+2048+4096)

A=8192-1

A=8191

                Đặt A = 1 + 2 + 4 + 8 + 16 +...+ 1024 + 2048 + 4096

                A = 1 + 2 + 2² + 2³ + 2⁴ +.... + 2¹⁰ + 2¹¹ + 2¹²

                     2A = 2 + 2² + 2³ + 2⁴ + 2⁵ + ..... + 2¹¹ + 2¹² + 2¹³

               2A - A = (2 + 2² + 2³ + 2⁴ + .... + 2¹¹ + 2¹² + 2¹³) - (1 + 2 + 2² + 2³ + 2⁴ +.....+ 2¹⁰ + 2¹¹ + 2¹²)

              => A = 2¹³ - 1

              => A = 8192 - 1 = 8191

           Ủng hộ mk nha !!! ^_^

= 128/256 + 64/256 + 32/256 + 16/256 + 8/256 + 4/256 + 2/256 + 1/256 

= 255/256

q​uy đ​ồ​ng mẫ​u số​ bạ​n nhé​. Mẫ​u số chung = 256

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