Ta có:

\(A=\dfrac{7\left(4-7^{2020}\right)}{7^{2021}}+\dfrac{5+7^{2021}}{7^{2021}}\)

\(A=\dfrac{28-7^{2021}+5+7^{2021}}{7^{2021}}=\dfrac{33}{7^{2021}}\)

Ta có: \(B=\dfrac{7^2}{7^{2021}}=\dfrac{49}{7^{2021}}\)

=> B>A

Ta có:

\(A=\frac{4-7^{2020}}{7^{2020}}+\frac{5+7^{2021}}{7^{2021}}\) và \(B=\frac{1}{7^{2019}}\)

Ta xét 2 trường hợp:

\(TH1:\frac{4-7^{2020}}{7^{2020}}=\frac{-7^{2020}+4}{7^{2020}}=-1+\frac{4}{7^{2020}}\)

\(TH2:\frac{5+7^{2021}}{7^{2021}}=1+\frac{5}{7^{2021}}\)

\(\Rightarrow\left(-1+\frac{4}{7^{2020}}\right)+\left(1+\frac{5}{7^{2021}}\right)\)

\(\Rightarrow\frac{4}{7^{2020}}+\frac{5}{7^{2021}}\)

\(Do:\)

\(\frac{4}{7^{2020}}>\frac{1}{7^{2019}}\)

\(\frac{5}{7^{2021}}>\frac{1}{7^{2019}}\)

Nên:\(\frac{4}{7^{2020}}+\frac{5}{7^{2021}}>\frac{1}{7^{2019}}\)

\(\Rightarrow A>B\)

1. \(\dfrac{2019}{2020}-\left(\dfrac{2019}{2020}-\dfrac{2020}{2021}\right)\)

\(=\dfrac{2019}{2020}-\dfrac{2019}{2020}+\dfrac{2020}{2021}\)

\(=0+\dfrac{2020}{2021}=\dfrac{2020}{2021}\)

Giải:

1) \(\dfrac{2019}{2020}-\left(\dfrac{2019}{2020}-\dfrac{2020}{2021}\right)\)  

\(=\dfrac{2019}{2020}-\dfrac{2019}{2020}+\dfrac{2020}{2021}\) 

\(=\left(\dfrac{2019}{2020}-\dfrac{2019}{2020}\right)+\dfrac{2020}{2021}\) 

\(=0+\dfrac{2020}{2021}\) 

\(=\dfrac{2020}{2021}\) 

2) \(\dfrac{2}{9}+\dfrac{7}{9}:\left(\dfrac{42}{5}-\dfrac{7}{5}\right)\) 

\(=\dfrac{2}{9}+\dfrac{7}{9}:7\) 

\(=\dfrac{2}{9}+\dfrac{1}{9}\) 

\(=\dfrac{1}{3}\) 

3) \(\dfrac{3}{4}+\dfrac{x}{4}=\dfrac{5}{8}\) 

            \(\dfrac{x}{4}=\dfrac{5}{8}-\dfrac{3}{4}\) 

            \(\dfrac{x}{4}=\dfrac{-1}{8}\)  

\(\Rightarrow x=\dfrac{4.-1}{8}=\dfrac{-1}{2}\) 

4) \(\left|3x+1\right|-\dfrac{1}{4}=\dfrac{-1}{4}\) 

            \(\left|3x-1\right|=\dfrac{-1}{4}+\dfrac{1}{4}\) 

            \(\left|3x-1\right|=0\) 

             \(3x-1=0\) 

                    \(3x=0+1\) 

                    \(3x=1\) 

                      \(x=1:3\) 

                      \(x=\dfrac{1}{3}\) 

Chúc bạn học tốt!

4) \(\left|3x+1\right|-\dfrac{1}{4}=\dfrac{-1}{4}\) 

            \(\left|3x+1\right|=\dfrac{-1}{4}+\dfrac{1}{4}\) 

            \(\left|3x+1\right|=0\) 

              \(3x+1=0\) 

                    \(3x=0-1\) 

                    \(3x=-1\) 

                      \(x=-1:3\) 

                      \(x=\dfrac{-1}{3}\)

Sửa đề :

1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + 9 - ... + 2018 - 2019 - 2020 + 2021

= 1 + ( 2 - 3 - 4 + 5 ) + ( 6 - 7 - 8 + 9 ) + ... + ( 2018 - 2019 - 2020 + 2021 )

= 1 + 0 + 0 + ... + 0

= 1

\(A=7^{2022}-7^{2021}+7^{2020}-7^{2019}+...+7^2-7\)

\(\Rightarrow7A=7^{2023}-7^{2022}+7^{2021}-...+7^3-7^2\)

\(\Rightarrow8A=A+7A=7^{2022}-7^{2021}+...+7^2-7+7^{2023}-7^{2022}+...+7^3-7^2=7^{2023}-7\)

\(\Rightarrow A=\dfrac{7^{2023}-7}{8}\)

Lời giải:
a.

$5+3(-7)+4:(-2)=5+(-21)+(-2)=5-(21+2)=5-23=-(23-5)=-18$

b.

$1-2-3+4+5-6-7+8+....+2017-2018-2019+2020+2021$

$=(1-2-3+4)+(5-6-7+8)+....+(2017-2018-2019+2020)+2021$

$=0+0+....+0+2021=2021$

S = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + ... + 2018 – 2019 - 2020 + 2021 + 2022

S = (1 + 2 - 3 - 4) + ... + (2017 + 2018 – 2019 - 2020) + 2021 + 2022

S = (-4) + ... + (-4) + 2021 + 2022

2020 : 4 = 505

S = (-4) . 505 + 2021 + 2022

S = (-2020) + 2021 + 2022

S = 2023

1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + ... + 2018 – 2019 - 2020 + 2021 

=(1 + 2 - 3 - 4) + ... + (2017 + 2018 – 2019 - 2020) + 2021 

= (-4) + ... + (-4) + 2021 + 2020 : 4 = 505 

= (-4) . 505 + 2021  

= (-2020) + 2021  

= 1

I'm come back:Đ

=(-1)+(-1)+...+(-1)+2021

=2021-1010

=1011

\(1-2+3-4+5-6+...+2019-2020+2021\)

\(=\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+...+\left(2019-2020\right)+2021\)

\(=-1-1-1-..-1+2021\)

\(=-1\cdot1010+2021\)

\(=-1010+2021\)

\(=-1011\)

$1-2+3-4+5-6+...+2019-2020+2021$

$=\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+...+\left(2019-2020\right)+2021$

$=-1-1-1-..-1+2021$

$=-1\cdot 1010+2021$

$=-1010+2021$

$=-1011$