2020/5021<2021/5023

bạn quy đồng lên sau đó làm nhé.

chúc bạn học tốt!

quy đòng lên bn dzợ

Ta có: 1+2-3-4+5+6-7-8+.....-2019-2020+2021+2022

=1+(2-3-4+5)+(6-7-8+9)+.....+(2018-2019-2020+2021)+2022

=1+0+0+.....+0+2022

=2023

\(A=1-3+5-7+......-2019+2021-2023\)

\(A=\left(1-3\right)+\left(5-7\right)+....+\left(2021-2023\right)\)

\(A=-2+\left(-2\right)+....+\left(-2\right)\left(506 cặp\right)\)

\(A=-2.506\)

\(A=-1012\)

*) A=(1-3)+(5-7)+....+(2021-2023)

<=> A=-2+(-2)+...+(-2)

Dãy A có (2023-1):2+1=1012 số số hạng 

=> Có 506 số (-2)

=> A=(-2).506=-1012

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

\(B=\left(1+2-3-4\right)+\left(5+6-7-8\right)+.....+\left(2017+2018-2019-2020\right)\)

\(B=-4+\left(-4\right)+.....+\left(-4\right)\left(505 cặp\right)\)

\(B=-4.505\)

\(B=-2020\)

\(\frac{x-4}{2021}+\frac{x-3}{2020}=\frac{x-2}{2019}+\frac{x-1}{2018}\)

\(\Leftrightarrow\left(\frac{x-4}{2021}+1\right)+\left(\frac{x-3}{2020}+1\right)=\left(\frac{x-2}{2019}+1\right)+\left(\frac{x-1}{2018}+1\right)\)

\(\Leftrightarrow\frac{x+2017}{2021}+\frac{x+2017}{2020}=\frac{x+2017}{2019}+\frac{x+2017}{2018}\)

\(\Leftrightarrow\frac{x+2017}{2021}+\frac{x+2017}{2020}-\frac{x+2017}{2019}-\frac{x+2017}{2018}=0\)

\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2021}+\frac{1}{2020}-\frac{1}{2019}-\frac{1}{2018}\right)=0\)

Mà \(\left(\frac{1}{2021}+\frac{1}{2020}-\frac{1}{2019}-\frac{1}{2018}\right)\ne0\)

\(\Leftrightarrow x+2017=0\)

\(\Leftrightarrow x=-2017\)

Vậy ..

=> (x-4/2021 +1) + (x-3/2020 +1) = (x-2/2019 +1)+ (x-1/2018 +1)

=> x+2017/2021 + x+2017/2020 = x+2017/2019 + x+2017/2018

=> x+2017/2018 + x+2017/2018 - x+2017/2020 - x+2017/2021 = 0

=> (x+2017).(1/2018+1/2019+1/2020+1/2021) = 0

=> x+2017 = 0 ( vì 1/2018+1/2019+1/2020+1/2021 > 0 )

=> x=-2017

Vậy x=-2017

k mk nha

\(\frac{x-4}{2021}+\frac{x-3}{2020}=\frac{x-2}{2019}+\frac{x-1}{2018}\)

\(\left(\frac{x-4}{2021}+1\right)+\left(\frac{x-3}{2020}+1\right)=\left(\frac{x-2}{2019}+1\right)+\left(\frac{x-1}{2018}+1\right)\)

\(\frac{x-2017}{2021}+\frac{x-2017}{2020}=\frac{x-2017}{2019}+\frac{x-2017}{2018}\)

\(\frac{x-2017}{2021}+\frac{x-2017}{2020}-\frac{x-2017}{2019}-\frac{x-2017}{2018}=0\)

\(\left(x-2017\right).\left(\frac{1}{2021}+\frac{1}{2020}-\frac{1}{2019}-\frac{1}{2018}\right)=0\)

vì \(\frac{1}{2021}+\frac{1}{2020}-\frac{1}{2019}-\frac{1}{2018}\ne0\)nên x - 2017 = 0 \(\Rightarrow\)x = 2017

A = (2020 : 4 + 20 : 4) x 2020 x 2021 x (3,9 - 2,6 x 1,5)

A = (2020 : 4 + 20 : 4) x 2020 x 2021 x 0

A = 0

Lời giải:
$A=1-\frac{1}{2019}+1-\frac{1}{2020}+1-\frac{1}{2021}+1+\frac{3}{2018}$

$=4+(\frac{1}{2018}-\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2020}+\frac{1}{2018}-\frac{1}{2021})$

$> 4+0+0+0+0=4$

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

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

            \(\left|3x+1\right|=0\) 

              \(3x+1=0\) 

                    \(3x=0-1\) 

                    \(3x=-1\) 

                      \(x=-1:3\) 

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