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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(4M=4^2+4^3+4^4+...+4^{2021}+4^{2022}\)
\(\Rightarrow4M-M=4^2+4^3+4^4+...+4^{2021}+4^{2022}-\left(4+4^2+4^3+...+4^{2020}+4^{2021}\right)\)
\(\Leftrightarrow3M=4^{2022}-4\)
\(\Leftrightarrow M=\dfrac{4^{2022}-4}{3}\)
\(4M=4^2+4^3+...+4^{2022}\)
\(\Leftrightarrow M=\dfrac{4^{2022}-4}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1. \(S=1+3+3^2+3^3+........+3^{2019}+3^{2020}\)
\(\Rightarrow3S=3+3^2+3^3+3^4+........+3^{2020}+3^{2021}\)
\(\Rightarrow3S-S=3^{2021}-1\)
\(\Rightarrow2S=3^{2021}-1\)
\(\Rightarrow S=\frac{3^{2021}-1}{2}\)
2. \(\left(3x-2\right)^3=64\)
\(\Leftrightarrow\left(3x-2\right)^3=4^3\)
\(\Leftrightarrow3x-2=4\)
\(\Leftrightarrow3x=6\)
\(\Leftrightarrow x=2\)
Vậy \(x=2\)
[3x-2]^3=64
Ta có:64=4^3
Suy ra:3x-2=4
3x =4+2
3x=6
x=6:3
x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
Các số lẻ trong khoảng từ 2000 đến 2020 đều là hợp số bởi vì những số đó có nhiều hơn hai ước
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)...\left(1-\frac{1}{2019}\right)\left(1-\frac{1}{2020}\right)\)
\(B=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot...\cdot\frac{2018}{2019}\cdot\frac{2019}{2020}\)
Số nào xuất hiện 2 lần thì thay thế những số đó bằng số 1.
\(B=\frac{1}{2020}\)
B = \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2019}\right).\left(1-\frac{1}{2020}\right)\)
= \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2018}{2019}.\frac{2019}{2020}\)
= \(\frac{1.2.3...2019}{2.3.4..2020}\)(Nếu có 2 thừa số giống nhau lặp lại ở tử số và mẫu số thì rút gọn coi như triệt tiêu hết và không có gì)
= \(\frac{1}{2020}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2T=2+\dfrac{3}{2^1}+\dfrac{4}{2^2}+...+\dfrac{2020}{2^{2018}}+\dfrac{2021}{2^{2019}}\)
\(T=2T-T=2+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2019}}-\dfrac{2021}{2^{2020}}\).
Đặt \(S=\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2019}}\Rightarrow2S=1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2018}}\Rightarrow S=2S-S=1-\dfrac{1}{2^{2019}}\).
Từ đó \(T=2+1-\dfrac{1}{2^{2019}}-\dfrac{2021}{2^{2020}}< 3\).