1/ \(\left(\dfrac{2021}{2020}+\dfrac{2020}{2021}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)

=\(\left(\dfrac{2021}{2020}+\dfrac{2020}{2021}\right).0\)

=\(0\)

mink chịu bài này nó rất khó

\(=\left(3-8+\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{4}{5}+\dfrac{1}{5}\right)-1\)

\(=-5-1-1=-7\)

a) Ta có:

Suy ra: 

Do đó .

Lại có .

Vậy A < B.

Sửa đề : \(M=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{1\dfrac{1}{6}-\dfrac{7}{6}+\dfrac{7}{10}}\right):\dfrac{2021}{2022}\)

\(M=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{1\dfrac{1}{6}-\dfrac{7}{6}+\dfrac{7}{10}}\right):\dfrac{2021}{2022}\\ =\left(\dfrac{2\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{7}{11}\right)}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{6}+\dfrac{7}{10}}\right):\dfrac{2021}{2022}\\ =\left(\dfrac{2}{7}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{\dfrac{7}{2}\left(\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}\right)}\right):\dfrac{2021}{2020}\\ =\left(\dfrac{2}{7}-\dfrac{2}{7}\right):\dfrac{2021}{2022}=0\)

a) \(A=2A-A\)

\(=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)\)

\(=1+\dfrac{1}{2}+...+\dfrac{1}{2^{2021}}-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)\)

\(=1-\dfrac{1}{2^{2022}}\)

b) \(B=\dfrac{20+15+12+17}{60}=\dfrac{4}{5}=1-\dfrac{1}{5}\)

\(A>B\left(Vì\left(\dfrac{1}{2^{2022}}< \dfrac{1}{5}\right)\right)\)

a) A = 2 A − A = 2 ( 1 2 + 1 2 2 + . . . + 1 2 2022 ) − ( 1 2 + 1 2 2 + . . . + 1 2 2022 ) = 1 + 1 2 + . . . + 1 2 2021 − ( 1 2 + 1 2 2 + . . . + 1 2 2022 ) = 1 − 1 2 2022 b) B = 20 + 15 + 12 + 17 60 = 4 5 = 1 − 1 5 A > B ( V ì ( 1 2 2022 < 1 5 ) )

1:

a: =7/5(40+1/4-25-1/4)-1/2021

=21-1/2021=42440/2021

b: =5/9*9-1*16/25=5-16/25=109/25

\(\dfrac{2}{5}\times\dfrac{3}{4}-\dfrac{1}{8}=\dfrac{1}{5}\times\dfrac{3}{2}-\dfrac{1}{8}=\dfrac{3}{10}-\dfrac{1}{8}=\dfrac{24}{80}-\dfrac{10}{80}=\dfrac{14}{80}=\dfrac{7}{40}\\ \dfrac{4}{3}+\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{5}{3}-\dfrac{1}{5}=\dfrac{25}{15}-\dfrac{3}{15}=\dfrac{22}{15}\\ \dfrac{9}{20}-\dfrac{3}{5}\times\dfrac{1}{4}=\dfrac{9}{20}-\dfrac{3}{20}=\dfrac{6}{20}=\dfrac{3}{10}\\ \dfrac{2}{8}+\dfrac{2}{3}:\dfrac{4}{5}=\dfrac{2}{8}+\dfrac{2}{3}\times\dfrac{5}{4}=\dfrac{2}{8}+\dfrac{1}{3}\times\dfrac{5}{2}=\dfrac{2}{8}+\dfrac{5}{6}=\dfrac{1}{4}+\dfrac{5}{6}=\dfrac{6}{24}+\dfrac{20}{24}=\dfrac{26}{24}=\dfrac{13}{12}\)

\(2021-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)\)

\(=\left(1-1\right)+\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+...+\left(1-\dfrac{1}{2021}\right)\)

\(=\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{2020}{2021}\)

Giải:

\(2021-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)=\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{2020}{2021}\) 

Ta có:

\(2021-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)\) 

\(=\left(1-1\right)+\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+...+\left(1-\dfrac{1}{2021}\right)\) 

\(=0+\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{2020}{2021}\) 

\(=\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{2020}{2021}\) 

Mà \(\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{2020}{2021}=\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{2020}{2021}\) 

\(\Rightarrow2021-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)=\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{2020}{2021}\left(đpcm\right)\)

\(A=1\dfrac{1}{2}\times1\dfrac{1}{3}\times1\dfrac{1}{4}\times....\times1\dfrac{1}{2020}\times1\dfrac{1}{2021}\\ =\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times\dfrac{6}{5}\times....\times\dfrac{2022}{2021}\\ =\dfrac{3\times4\times5\times6\times.....\times2022}{2\times3\times4\times5\times....\times2021}\\ =\dfrac{2022}{2}=1011\)