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

\(\Rightarrow A>B\)

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

bài 1:

ssh của A là:

(151-3):2+1=75

A=(151+3)x75:2=5775

đáp số: 5775

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

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

            \(\left|3x+1\right|=0\) 

              \(3x+1=0\) 

                    \(3x=0-1\) 

                    \(3x=-1\) 

                      \(x=-1:3\) 

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

a) 29^57 < 29^75

b) 1011^22 < 1101^22

c) ( 2021 + 2018 )^2019 < ( 2020 + 2019 )^2020

d) 2^5000 > 7^2000

Chúc bạn học tốt @!!!

Nếu có thể thì t.i.c.k cho mình nha ! Thank

Đại ca toàn học giỏi nhất hành tinh đây

a) 29⁵⁷và 29⁷⁵

 ^=> 29⁵⁷<29⁷⁵

B=-1.1009+2019+2020+2021

a) Số số hạng=(2020-5):5+1=404

=> A=(2020+5)×404 : 2= 409050

b) Từ 1 đến 2020 có 2020 số hạng=> có 1010 cặp,mỗi cặp = -1

=>B= 1010×(-1) +2021=1011

a) =[(2020-5):5+1].[(2020+5):5]

=403.405

=163215

SSHẠNG :(2020-5) : 5+1=404 ( 2021-1)+1=2021

= (2020+5). 404:2 (2021+1) :2.2021

A= 409050 B= 2043231

Lời giải:
a.

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

b.

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

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

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