A = 2⁵.(-5)² - 8² - 7

= 32.25 - 64 - 7

= 729

= 27²

B = 2³.(-4)² + (-3)².3² - 40

= 8.16 + 9.9 - 40

= 169

= 13²

C = (1/4 - 1/2 - 1)³ . (2 - 2/5)³

= (-5/4)³ . (8/5)³

= (-5/4 . 8/5)³

= (-2)³

D = (-1/4)² : (1/2 - 1/3)

= 1/16 : 1/6

= 3/8

E = 4 . (1/4)² + 25 . [(3/4)³ : (5/4)³] : (3/2)³

= 1/4 + 25 . (3/4 . 5/4)³ : (3/2)³

= 1/4 + 25 . (15/16)³ : 27/8

= 1/4 + 25 . 3375/4096 : 27/8

= 1/4 + 84375/4096 : 27/8

= 1/4 + 3125/512

= 3253/512

F = 2³ + 3.(1/2)⁰ - 1 + [(-2)² : 1/2] - 8

= 8 + 3.1 - 1 + (4 : 1/2) - 8

= 8 + 3 - 1 + 8 - 8

= 10

Lời giải:
Gọi tổng trên là $A$
$A=\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+....+\frac{1}{\frac{2023.2024}{2}}$

$=\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2023.2024}$

$=2(\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{2024-2023}{2023.2024})$

$=2(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{2023}-\frac{1}{2024})$

$=2(\frac{1}{3}-\frac{1}{2024})=\frac{2021}{3036}$

$A=\genfrac{}{}{0ex}{}{1}{\genfrac{}{}{0ex}{}{3.4}{2}}+\genfrac{}{}{0ex}{}{1}{\genfrac{}{}{0ex}{}{4.5}{2}}+....+\genfrac{}{}{0ex}{}{1}{\genfrac{}{}{0ex}{}{2023.2024}{2}}$

$=\genfrac{}{}{0ex}{}{2}{3.4}+\genfrac{}{}{0ex}{}{2}{4.5}+...+\genfrac{}{}{0ex}{}{2}{2023.2024}$

$=2(\genfrac{}{}{0ex}{}{4-3}{3.4}+\genfrac{}{}{0ex}{}{5-4}{4.5}+...+\genfrac{}{}{0ex}{}{2024-2023}{2023.2024})$

$=2(\genfrac{}{}{0ex}{}{1}{3}-\genfrac{}{}{0ex}{}{1}{4}+\genfrac{}{}{0ex}{}{1}{4}-\genfrac{}{}{0ex}{}{1}{5}+....+\genfrac{}{}{0ex}{}{1}{2023}-\genfrac{}{}{0ex}{}{1}{2024})$

$=2(\genfrac{}{}{0ex}{}{1}{3}-\genfrac{}{}{0ex}{}{1}{2024})=\genfrac{}{}{0ex}{}{2021}{3036}$
 

\(1-\frac{1}{2}+2-\frac{2}{3}+3-\frac{3}{4}+4-\frac{1}{4}-3-\frac{1}{3}-2-\frac{1}{2}-1\)

\(=\left(1+2+3+4-3-2-1\right)+\left(-\frac{1}{2}-\frac{2}{3}-\frac{3}{4}-\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)\)

\(=4+\left[\left(-\frac{1}{2}-\frac{1}{2}\right)+\left(-\frac{2}{3}-\frac{1}{3}\right)+\left(-\frac{3}{4}-\frac{1}{4}\right)\right]\)

\(=4+\left[\left(-1\right)+\left(-1\right)+\left(-1\right)\right]\)

\(=4+\left(-3\right)=1\)

\(1-\frac{1}{2}+2-\frac{2}{3}+3-\frac{3}{4}+4-\frac{1}{4}-3-\frac{1}{3}-2-\frac{1}{2}-1\)

\(=\left(1-1\right)+\left(2-2\right)+\left(3-3\right)+4-\left(\frac{1}{2}+\frac{1}{2}\right)-\left(\frac{2}{3}+\frac{1}{3}\right)-\left(\frac{3}{4}+\frac{1}{4}\right)\)

\(=4-1-1-1\)

\(=1\)