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![](https://rs.olm.vn/images/avt/0.png?1311)
=(1-2)-(3-4)+(5-6)-(7-8)+...+(2021-2022)-2023
=(-1)-(-1)+(-1)-...+(-1)-2023
=0-2023
=-2023
![](https://rs.olm.vn/images/avt/0.png?1311)
A = \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\)
= \(1-\dfrac{1}{2023}\)
= \(\dfrac{2022}{2023}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`3/4-(2/3+3/4)+2/3+2022/2023`
`=3/4 - 2/3 - 3/4 +2/3 +2022/2023`
`= (3/4 -3/4 ) + (-2/3 +2/3) +2022/2023`
`= 0+0+2022/2023`
`=2022/2023`
\(\dfrac{3}{4}-\left(\dfrac{2}{3}+\dfrac{3}{4}\right)+\dfrac{2}{3}+\dfrac{2022}{2023}\)
\(=\dfrac{3}{4}-\left(\dfrac{8}{12}+\dfrac{9}{12}\right)+\dfrac{2}{3}+\dfrac{2022}{2023}\)
\(=\dfrac{3}{4}-\dfrac{17}{12}+\dfrac{2}{3}+\dfrac{2022}{2023}\)
\(=\dfrac{9}{12}-\dfrac{17}{12}+\dfrac{8}{12}+\dfrac{2022}{2023}\)
\(=\dfrac{9-17+8}{12}+\dfrac{2022}{2023}=\dfrac{0}{12}+\dfrac{2022}{2023}=0+\dfrac{2022}{2023}\)
\(=\dfrac{2022}{2023}\)
#YTVA
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=1-3+3^2-3^3+...+3^{2021}-3^{2022}\)
\(3A=3-3^2+3^3-3^4+...+3^{2022}-3^{2023}\)
\(3A-A=\left(1-3+3^2-3^3+...+3^{2021}-3^{2022}\right)-\left(3-3^2+3^3-3^4+...+3^{2022}-3^{2023}\right)\)
\(2A=3^{2023}-1\)
\(\Rightarrow A=\left(3^{2023}-1\right)\div2\)
\(\text{cái này mình sợ sai nên bạn có thể nhờ cô chữa}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`2x-15=-25`
`2x=-10`
`x=-5`
___________
`3/5<x/10<4/5`
`3/5=(3xx10)/(5xx10)=30/50`
`x/10=(5x)/(10xx5)=(5x)/50`
`4/5=(4xx10)/(5xx10)=40/50`
`=>30/50<(5x)/50<40/50`
`=>30<5x<40`
`=>x=7`
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn nên viết lại đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề của bạn hơn nhé.
![](https://rs.olm.vn/images/avt/0.png?1311)
1: \(A=6^{2020}\left(1+6\right)+6^{2022}\left(1+6\right)\)
\(=7\left(6^{2020}+6^{2022}\right)⋮7\)
Bài 1:
$A=6^{2020}(1+6+6^2+6^3)=6^{2020}.259=6^{2020}.7.37\vdots 7$
Ta có đpcm.
\(A=1\cdot2+2\cdot3+...+n\left(n+1\right)\\ \Rightarrow3\cdot A=1\cdot2\cdot3+2\cdot3\cdot3+...+n\left(n+1\right)\cdot3\)
\(\Rightarrow3\cdot A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+n\left(n+1\right)\left(n+2-\left(n-1\right)\right)\)\(\Rightarrow3\cdot A=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+3\cdot4\cdot5-2\cdot3\cdot4+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)\(\Rightarrow3\cdot A=n\left(n+1\right)\left(n+2\right)\)
\(\Rightarrow A=\dfrac{n\left(n+1\right)\left(n+2\right)}{3}\)