\(\dfrac{3}{4}\) + \(\dfrac{3}{4}\) = \(\dfrac{6}{4}\) = \(\dfrac{3}{2}\)

\(\dfrac{3}{4}+\dfrac{3}{4}\\ =\dfrac{3+3}{4}\\ =\dfrac{6}{4}\\ =\dfrac{6:2}{4:2}\\ =\dfrac{3}{2}\)

\(\dfrac{2021\times2023-1}{2020\times2023+2022}\\ =\dfrac{2023\times\left(2020+1\right)-1}{2023\times2020+2022}\\ =\dfrac{2023\times2020+2023\times1-1}{2023\times2020+2022}\\ =\dfrac{2023\times2020+2023-1}{2023\times2020+2022}\\ =\dfrac{2023\times2020+\left(2023-1\right)}{2023\times2020+2022}\\ =\dfrac{2023\times2020+2022}{2023\times2020+2022}\\ =1\)

\(\dfrac{2021\times2023-1}{2020\times2023+2022}=\dfrac{\left(2020+1\right)\times2023-1}{2020\times2023+2022}=\dfrac{2020\times2023+1\times2023-1}{2020\times2023+2022}\)

\(=\dfrac{2020\times2023+2022}{2020\times2023+2022}=1\)

\(\dfrac{1}{3^2}< \dfrac{1}{1\cdot3}=1-\dfrac{1}{3}\)

\(\dfrac{1}{5^2}< \dfrac{1}{3\cdot5}=\dfrac{1}{3}-\dfrac{1}{5}\)

...

\(\dfrac{1}{99^2}< \dfrac{1}{97\cdot99}=\dfrac{1}{97}-\dfrac{1}{99}\)

Do đó: \(\dfrac{1}{3^2}+\dfrac{1}{5^2}+...+\dfrac{1}{99^2}< 1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\)

=>\(B=1+\dfrac{1}{3^2}+\dfrac{1}{5^2}+...+\dfrac{1}{9^2}< 1+1=2\)

=>1<B<2

=>B không là số tự nhiên

Số các số hạng của A:

(133 - 1) : 12 + 1 = 12 (số)

Tổng các tử số:

(133 + 1) × 12 : 2 = 804

A = 804/1000 = 201/250

12,1 > 12 > 11 > 31/3

Số bé nhất thỏa mãn là 11

Vậy x = 11

12,1 > \(x\) > \(\dfrac{32}{3}\)

12\(\dfrac{1}{10}\) > \(x\) > 10\(\dfrac{2}{3}\)

Vậy \(x\) = 11

\(\dfrac{10\dfrac{1}{3}\text{x}\left(26\dfrac{1}{7}-\dfrac{176}{7}\right)-\dfrac{12}{11}\text{x}\left(\dfrac{10}{3}-1,75\right)}{\left(\dfrac{5}{9}-0,25\right)\text{x}\dfrac{60}{11}-1}\)

\(=\dfrac{\dfrac{31}{3}\text{x}\left(\dfrac{183}{7}-\dfrac{176}{7}\right)-\dfrac{12}{11}\text{x}\left(\dfrac{10}{3}-\dfrac{7}{4}\right)}{\left(\dfrac{5}{9}-\dfrac{1}{4}\right)\text{x}\dfrac{60}{11}-1}\)

\(=\dfrac{\dfrac{31}{3}-\dfrac{12}{11}\text{x}\dfrac{19}{12}}{\dfrac{20-9}{36}\text{x}\dfrac{60}{11}-1}==\dfrac{\dfrac{31}{3}-\dfrac{19}{11}}{\dfrac{5}{3}-1}\)

\(=\dfrac{284}{33}:\dfrac{2}{3}=\dfrac{284}{33}\text{x}\dfrac{3}{2}=\dfrac{142}{11}\)

Hiệu ban đầu của hai số là:

20,15-3,6-2,5=20,15-6,1=14,05

Số lớn là (43,75+14,05):2=28,9

Số bé là 28,9-14,05=14,85

   (\(x\) - 1) - (y - 1) 

= \(x\) - 1 - y + 1

= (\(x\) - y)  - (1 - 1)

= \(x\) - y - 0

= \(x\) - y

Vậy phép tính trên là đúng.

Phép tính trên đúng