Tính tổng
S=1+5+13+...+2017+2021+2022+2023
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) A=6 -13 +(-14+15+16-17)+(-18+19+20-21)+...+(-2018+2019+2020-2021)+(-2022+2023+2024-2025) +2025
A=-7 +0 +0 +...+0+0 +2025= 2018
B) 7-9+(-10+11+12-13)+(-14+15+16-17)+...+(-2018+2019+2020-2021)+2021
B= -2+0+0+...+0+2021=2019
#Có gì không hiểu thì hỏi nha#
\(\dfrac{2022\times2023-1}{2023\times2021+2022}\)
= \(\dfrac{\left(2021+1\right)\times2023-1}{2023\times2021+2022}\)
= \(\dfrac{2023\times2021+2023-1}{2023\times2021+2022}\)
= \(\dfrac{2023\times2021+2022}{2023\times2021+2022}\)
= 1
2017/2020<2019/2020< 1
1< 2022/2021< 2023/2021
vậy phân số lớn nhất là 2023/2021
ta so sánh với 1:
2017/2020<2019/2020< 1
1< 2022/2021< 2023/2021
nên phân số lớn nhất là phân số cuối: 2023/2021
a) A=6 -13 +(-14+15+16-17)+(-18+19+20-21)+...+(-2018+2019+2020-2021)+(-2022+2023+2024-2025) +2025
A=-7 +0 +0 +...+0+0 +2025= 2018
B) 7-9+(-10+11+12-13)+(-14+15+16-17)+...+(-2018+2019+2020-2021)+2021
B= -2+0+0+...+0+2021=2019
#Có gì không hiểu thì hỏi nha#
1-3-5+7+9-11-13+15+...+2017-2019-2021+2023=
=(1-3-5+7)+(9-11-13+15)+...+(2017-2019-2021+2023)=
=0+0+.....+0=0
\(\frac{x+1}{2019}+\frac{x+2}{2018}+\frac{x+3}{2017}=\frac{x-1}{2021}+\frac{x-2}{2022}+\frac{x-3}{2023}\)
\(\Leftrightarrow\left(\frac{x+1}{2019}+1\right)+\left(\frac{x+2}{2018}+1\right)+\left(\frac{x+3}{2017}+1\right)=\left(\frac{x-1}{2021}+1\right)+\left(\frac{x-2}{2022}+1\right)+\left(\frac{x-3}{2023}+1\right)\)
\(\Leftrightarrow\left(\frac{x+1+2019}{2019}\right)+\left(\frac{x+2+2018}{2018}\right)+\left(\frac{x+3+2017}{2017}\right)=\left(\frac{x-1+2021}{2021}\right)+\left(\frac{x-2+2022}{2022}\right)+\left(\frac{x-3+2023}{2023}\right)\)
\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}=\frac{x+2020}{2021}+\frac{x+2020}{2022}+\frac{x+2020}{2023}\)
\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}-\frac{x+2020}{2021}-\frac{x+2020}{2022}-\frac{x+2020}{2023}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2021}-\frac{1}{2022}-\frac{1}{2023}\right)=0\)
Vì \(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2021}-\frac{1}{2022}-\frac{1}{2023}\ne0\)
=> x + 2020 = 0
=> x = -2020
Bài làm :
Ta có :
\(\frac{x+1}{2019}+\frac{x+2}{2018}+\frac{x+3}{2017}=\frac{x-1}{2021}+\frac{x-2}{2022}+\frac{x-3}{2023}\)
\(\Leftrightarrow\left(\frac{x+1}{2019}+1\right)+\left(\frac{x+2}{2018}+1\right)+\left(\frac{x+3}{2017}+1\right)=\left(\frac{x-1}{2021}+1\right)+\left(\frac{x-2}{2022}+1\right)+\left(\frac{x-3}{2023}+1\right)\)
\(\Leftrightarrow\left(\frac{x+1+2019}{2019}\right)+\left(\frac{x+2+2018}{2018}\right)+\left(\frac{x+3+2017}{2017}\right)=\left(\frac{x-1+2021}{2021}\right)+\left(\frac{x-2+2022}{2022}\right)+\left(\frac{x-3+2023}{2023}\right)\)
\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}=\frac{x+2020}{2021}+\frac{x+2020}{2022}+\frac{x+2020}{2023}\)
\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}-\frac{x+2020}{2021}-\frac{x+2020}{2022}-\frac{x+2020}{2023}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2021}-\frac{1}{2022}-\frac{1}{2023}\right)=0\)
\(\text{Vì : }\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2021}-\frac{1}{2022}-\frac{1}{2023}\ne0\)
\(\Rightarrow x+2020=0\Leftrightarrow x=-2020\)
Vậy x=-2020
=(1-2-3+4)+(5-6-7+8)+...+(2017-2018-2019+2020)+2021-2022-2023
=0+0+...+0-1-2023
=-2024
giúp mình với
Đặt S = B + 2022 + 2023
Số số hạng của B là : ( 2021 - 1 ) : 4 + 1 = 506 ( số )
Tổng B là : ( 2021 + 1 ) . 506 : 2 = 511566
=> S = 511566 + 2022 + 2023
=> S = 515611
Vậy,......