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Bài làm :
Ta có :
\(x+2x+3x+...+2020x=2020.2021\)
\(\Leftrightarrow x\left(1+2+3+...+2020\right)=2020.2021\)
\(\Leftrightarrow x.\frac{\left(2020+1\right).2020}{2}=2021.2020\)
\(\Leftrightarrow x.\frac{2021.2020}{2}=2021.2020\)
\(\Leftrightarrow x=2\)
Vậy x=2
\(x+2x+3x+...+2020x=2020\cdot2021\)
\(x\left(1+2+3+...+2020\right)=2020\cdot2021\)
1 + 2 + 3 ... + 2020
Số số hạng :
\(\left(2020-1\right):1+1=2020\)
Tổng :
\(\left(2020+1\right)\cdot2020:2=2021\cdot1010\)
\(2021\cdot1010\cdot x=2020\cdot2021\)
\(1010x=2020\)
\(x=2\)

K=2.(2/2.4+2/4.6+2/6.8+...+2/2008.2010)
K=2.(4-2/2.4+6-4/4.6+8-6/6.8+...+2010-2008/2008.2010)
K=2.(1/2-1/4+1/4-1/6+1/6-1/8+...+1/2008-1/2010)
K=2.(1.2-1.2010)
K=2.502/1005
K=1004/1005

\(\frac{4}{4.6}+\frac{4}{6.8}+\frac{4}{8.10}+...+\frac{4}{28.30}\)
\(=2.\left(\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+...+\frac{2}{28.10}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+...+\frac{1}{28}-\frac{1}{30}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{30}\right)=2.\left(\frac{15}{60}-\frac{2}{60}\right)=2.\frac{13}{60}=\frac{26}{60}=\frac{13}{30}\)

F=4/2.4+4/4.6+4/6.8+..........+4/2008.2010
F=2/2-2/4+2/4-2/6+2/6-2/8+......+2/2008-2/2010
F=2/2- 2/4+2/4-2/6+2/6-2/8+......+2/2008-2/2010
F=2/2-2/2010
=>F=2008/2010=1004/1005

Gọi A= 4/2.4+4/4.6+4/6.8+...+4/2008.2010
A/2= 2/2.4+2/4.6+...+2/2008.2010
Mà 2/2.4=1/2-1/4; 2/4.6=1/4-1/6 ....
Vậy A/2= (1/2-1/4)+(1/4-1/6)+....+(1/2008-1/2010)
A/2=1/2-1/2010=2010/4020-2/4020=2008/4...
A= 2008.2/4020=1004/1005
C = 4/2.4 + 4/4.6 + 4/6.8 + ... + 4/2008.2010
C = 2 . (2/2.4 + 2/4.6 + 2/6.8 + ... + 2/2008.2010)
C = 2 . (1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + ... + 1/2008 - 1/2010)
C = 2 . (1/2 - 1/2010)
C = 2 . 502/1005
C = 1004/1005

\(F=\frac{4}{2.4}+\frac{4}{4.6}+....+\frac{4}{2008.2010}\)
\(F=\frac{4}{2}.\left(\frac{1}{2.4}+\frac{1}{4.6}+....+\frac{1}{2008.2010}\right)\)
\(F=\frac{4}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(F=\frac{4}{2}.\left(\frac{1}{2}-\frac{1}{2010}\right)=\frac{4}{2}.\frac{502}{1005}=\frac{1004}{1005}\)
F=4/2.4+4/4.6+...+4/2008.2010
=2(2/2.4+2/4.6+....+2/2008.2010)
=2(1/2-1/4+1/4-1/6+....+1/2008-1/2010)
=2(1/2-1/2010)
=2.\(\frac{502}{1005}\)
=\(\frac{8032}{1005}\)

A = 4/2 x 4 + 4/4 x 6 + ..... + 4/2008 x 2010
A = 4/2 - 4/4 + 4/4 - 4/6 + ..... + 4/2008 - 4/2010
A = 4/2 - 4/2010
A = Dư bạn ơi !!!!!!!!!!!!!!!!!!

A=4/2.4+4/4.6+4/6.8+...+4/2008.2010
=2.(2/2.4+2/4.6+2/6.8+...+2/2008.2010)
=2.(1/2-1/4+1/4-1/6+1/6-1/8+...+1/2008-1/2010)
=2.(1/2-1/2010)
=2.502/1005
=1004/1005
Vậy A=1004/1005
100% giải đúng đầu tiên:
Ta có: \(A=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(=2.\frac{2}{2.4}+2.\frac{2}{4.6}+2.\frac{2}{6.8}+...+2.\frac{2}{2008.2010}\)
\(=2\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+..+\frac{2}{2008.2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(=2.\frac{1}{2}-2.\frac{1}{2010}\)
\(=1-\frac{1}{1005}=\frac{1004}{1005}\)
Trả lời:
Sai quy luật rồi bn ơi, bn chỉnh lại đề đi
~HT~
Có phải thế này ko:
Giải thích các bước giải:
4 / 2 . 4 + 4/ 4 . 6 +...........................+ 4/ 2020 . 2021
= 2 x ( 2 / 2 . 4 + 2 / 4 . 6 + ................... + 2 / 2020 . 2021
= 2 x ( 1/2 - 1/4 + 1/4 - 1/6 + 1/6 -1/8 + .............................+ 1/2020 - 1/2021 )
= 2 x ( 1/2 - 1/2021 )
= 2 x 2019/ 4042
= 2019/2021
Mk viết hơi khó nhìn mong bn thông cảm.