4/4.4+4/4.6+...+4/2020.2021
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![](https://rs.olm.vn/images/avt/0.png?1311)
4.4
TH1: Nếu x<1 thì ta có:
1 - x + 4 - x =3x
5-2x=3x
5=5x
=> x=1 ( k thỏa mãn)
TH2: 1<x<4 thì ta có :
x-1=4x=3x
3=3x
=> x=1 ( thỏa mãn)
TH3:Nếu x >4 thì ta có:
x-1+x-4=3x
2x-5=3x
2x-3x=5
-x=5
=> x=-5 ( k thỏa mãn)
Vậy x=1
![](https://rs.olm.vn/images/avt/0.png?1311)
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 :
\(\left(2020+1\right)\cdot2020:2=2021\cdot1010\)
\(2021\cdot1010\cdot x=2020\cdot2021\)
\(1010x=2020\)
\(x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{4}{5}-\dfrac{3}{4}.\dfrac{4}{5}.\dfrac{3}{4}\)
\(=\dfrac{4}{5}-\dfrac{9}{20}\)
\(=\dfrac{16}{20}-\dfrac{9}{20}\)
\(=\dfrac{7}{20}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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Sửa đề: \(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2018\cdot2020}+\dfrac{4}{2020\cdot2022}\)
Ta có: \(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2018\cdot2020}+\dfrac{4}{2020\cdot2022}\)
\(=2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2018\cdot2020}+\dfrac{2}{2020\cdot2022}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2018}-\dfrac{1}{2020}+\dfrac{1}{2020}-\dfrac{1}{2022}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{2022}\right)\)
\(=2\cdot\dfrac{505}{1011}\)
\(=\dfrac{1010}{1011}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{4}{4.6}+\frac{4}{6.8}+\frac{4}{8.10}+...+\frac{4}{28.30}\)
\(=2.\left(\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+...+\frac{2}{28.10}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+...+\frac{1}{28}-\frac{1}{30}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{30}\right)=2.\left(\frac{15}{60}-\frac{2}{60}\right)=2.\frac{13}{60}=\frac{26}{60}=\frac{13}{30}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
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.