Ta có: 
Mẫu số chung 2 phân số: 84
\(\dfrac{3}{7}=\dfrac{3*12}{7*12}=\dfrac{36}{84}\)
\(\dfrac{5}{12}=\dfrac{5*7}{12*7}=\dfrac{35}{84}\)
Vì \(36>35\) nên\(\dfrac{36}{84}>\dfrac{35}{84}\)
Vậy \(\dfrac{3}{7}>\dfrac{5}{12}\)

Ta có:

\(\dfrac{9}{8}>1>\dfrac{2023}{2024}\) nên \(\dfrac{9}{8}>\dfrac{2023}{2024}\)

Ta có:

\(\dfrac{1+15}{16}=1\)

\(\dfrac{1+16}{15}=\dfrac{17}{15}>1\)

\(\Rightarrow\dfrac{1+15}{16}>\dfrac{1+16}{15}\)

3/7 < 5/12

9/8 > 2023/2024

\(D=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{2048}\) (sửa đề)

\(\dfrac{1}{2}\cdot D=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+...+\dfrac{1}{4096}\)

\(D-\dfrac{1}{2}D=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{2048}-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+...+\dfrac{1}{4096}\right)\)

\(\dfrac{1}{2}D=1-\dfrac{1}{4096}\)

\(\dfrac{1}{2}D=\dfrac{4095}{4096}\)

\(\Rightarrow D=\dfrac{4095}{4096}:\dfrac{1}{2}=\dfrac{4095}{2048}\)

Vậy \(D=\dfrac{4095}{2048}\)

Đặt A = 1 + 2 + 3 + 4 + ... + 2023

Tổng có 2023 - 1 + 1 số hạng

A = (2023 + 1) × 2023 : 2

= 2047276

-----------------------

Đặt B = 20 + 21 + 22 + ... + 2024

Tổng có: 2024 - 20 + 1 = 2005 số hạng

B = (2024 + 20) × 2005 : 2

= 2049110

------------------------

Đặt C = 2 + 4 + 6 + ... + 2024

Tổng có (2024 - 2) : 2 + 1 = 1012 số hạng

C = (2024 + 2) × 1012 : 2

= 1025156

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Đặt D = 1 + 2 + 4 + 8 + 16 + ... + 8192

2 × D = 2 + 4 + 8 + 16 + 32 + ... + 16384

2 × D - D = (2 + 4 + 8 + 16 + 32 + ... + 16384) - (1 + 2 + 4 + 8 + 16 + ... + 8192)

= 16384 - 1

= 16383

Vậy D = 16383

\(a,A=1+2+3+4+5..+2023\)

Số số hạng:

\(\left(2023-1\right):1+1=2023\)

Tổng :

\(\dfrac{\left(2023+1\right).2023}{2}=2047276\)

\(b,20+21+22+..+2024\)

Số số hạng:

\(\left(2024-20\right):1+1=2005\)

Tổng:

\(\dfrac{\left(2024+20\right).2005}{2}=2049110\)

\(c,2+4+6+..+2024\)

Số số hạng:

\(\left(2024-2\right):2+1=1012\)

Tổng:

\(\dfrac{\left(2024+2\right).1012}{2}=1025156\)

1+2+3+4 +...+2023

Số phần tử là :

( 2023 -1 ) : 1 + 1 = 2023 ( phần tử )

Tổng các số là :

 ( 2023 + 1 ) x 2023 : 2 = 2047276 

20+21+22+...+2024

Số phần tử là :

(  2024 - 20 ) : 1 + 1 = 2005 ( phần tử ) 

Tổng các số là :

 ( 2024 + 20 ) x 2005 : 2 = 2049110 

2+4+6+...+2024

Số phần tử là :

( 2024 - 2 ) : 2 + 1 = 1012 ( phần tử )

Tổng các số là :

( 2024 + 2 ) x 1012 : 2 = 1025156 

1+2+4+8+16+....+ 8192

A = 2^0 + 2^1 + 2^2 + 2^3 +....+2^13

2A = 2^1 + 2^2 + 2^3 + 2^4 +..... + 2^14

2A - A = 2^14 - 2^0

=> A = 2^14 -1 

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

đặt A=(1/2+1/2^2+1/2^3+.....+1/2^10)

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

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

A=1-1/2^10

=-1-1-1/2^10=......tự làm nha

Đề chắc sai e ạ, a sửa luôn :

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

\(A=\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{10}}\)

\(2A=1-\frac{1}{2}-...-\frac{1}{2^9}\)

\(2A-A=\left(1-\frac{1}{2}-...-\frac{1}{2^9}\right)-\left(\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{10}}\right)\)

\(A=1-\frac{1}{2}-...-\frac{1}{2^9}-\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\)

\(A=1-\left(\frac{1}{2}+\frac{1}{2}\right)+\frac{1}{2^{10}}\)

\(A=\frac{1}{2^{10}}\)

ok thank

\(A=\dfrac{9}{8}-\dfrac{8}{9}+\dfrac{3}{24}+\dfrac{1}{4}-\dfrac{5}{16}+\dfrac{19}{25}-\dfrac{1}{9}+\dfrac{2}{25}-\dfrac{1}{81}\)

\(=\dfrac{9}{8}+\dfrac{1}{4}-\dfrac{5}{16}+\dfrac{1}{8}-\dfrac{8}{9}-\dfrac{1}{9}-\dfrac{1}{81}+\dfrac{19}{25}+\dfrac{2}{25}\)

\(=\dfrac{10}{8}+\dfrac{1}{4}-\dfrac{5}{16}-1-\dfrac{1}{81}+\dfrac{21}{25}\)

\(=\dfrac{20+4-5}{16}-\dfrac{82}{81}+\dfrac{21}{25}\)

\(=\dfrac{19}{16}-\dfrac{82}{81}+\dfrac{21}{25}\)

\(=\dfrac{32891}{16\cdot81\cdot25}\)

b: \(B=-\dfrac{1}{3}-\dfrac{8}{35}-\dfrac{2}{9}-\dfrac{1}{35}+\dfrac{4}{5}-\dfrac{4}{9}+\dfrac{3}{7}\)

\(=\dfrac{-1}{3}-\dfrac{2}{9}-\dfrac{4}{9}-\dfrac{8}{35}-\dfrac{1}{35}+\dfrac{4}{5}+\dfrac{3}{7}\)

\(=\dfrac{-3-2-4}{9}+\dfrac{-9}{35}+\dfrac{28+15}{35}\)

\(=-1+\dfrac{-9+43}{35}=-1+\dfrac{34}{35}=-\dfrac{1}{35}\)

a: =7-2+5=10