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Giả sử tất cả các số đã cho đều lẻ
=>Quy đồng, ta được:
\(A=\dfrac{\left(a_2\cdot a_3\cdot...\cdot a_{2022}\right)+\left(a_1\cdot a_3\cdot...\cdot a_{2021}\cdot a_{2022}\right)+...+\left(a_1\cdot a_2\cdot...\cdot a_{2021}\right)}{a_1\cdot a_2\cdot...\cdot a_{2022}}=1\)
Tử có 2022 số hạng, mẫu là số lẻ
=>A là số chẵn khác 1
=>Trái GT
=>Phải có ít nhất 1 số là số chẵn
a) 2021 - (1/3)² . 3²
= 2021 - 1/9 . 9
= 2021 - 1
= 2020
b) 5/10 + 9 . (-3/2)
= 1/2 - 27/2
= -26/2
= -13
c) -10 . (-2021/2022)⁰ + (2/5)² : 2
= -10 . 1 + 4/25 . 2
= -10 + 8/25
= -68/7
\(a,2021-\left(\dfrac{1}{3}\right)^2\cdot3^2\\ =2021-\dfrac{1}{9}\cdot9\\ =2021-\dfrac{9}{9}\\ =2021-1=2020\\ b,\dfrac{5}{10}+9\cdot\dfrac{-3}{2}\\ =\dfrac{5}{10}+\dfrac{-27}{2}\\ =\dfrac{5}{10}+\dfrac{-135}{10}\\ =-\dfrac{130}{10}\\ =-13\\ c,-10\cdot\left(-\dfrac{2021}{2022}\right)^0+\left(\dfrac{2}{5}\right)^2:2\\ =-10\cdot1+\dfrac{4}{25}\cdot\dfrac{1}{2}\\ =-10+\dfrac{4}{50}\\ =-10+\dfrac{2}{25}\\ =-\dfrac{248}{25}\)
a)
`(2x-1)(x+2/3)=0`
\(< =>\left[{}\begin{matrix}2x-1=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
b)
\(\dfrac{x+4}{2019}+\dfrac{x+3}{2020}=\dfrac{x+2}{2021}+\dfrac{x+1}{2022}\)
\(< =>\dfrac{x+4}{2019}+1+\dfrac{x+3}{2020}+1=\dfrac{x+2}{2021}+1+\dfrac{x+1}{2022}+1\)
\(< =>\dfrac{x+2023}{2019}+\dfrac{x+2023}{2020}=\dfrac{x+2023}{2021}+\dfrac{x+2023}{2022}\)
\(< =>\left(x+2023\right)\left(\dfrac{1}{2019}+\dfrac{1}{2020}-\dfrac{1}{2021}-\dfrac{1}{2022}\right)=0\)
\(< =>x+2023=0\left(\dfrac{1}{2019}+\dfrac{1}{2020}-\dfrac{1}{2021}-\dfrac{1}{2022}\ne0\right)\\ < =>x=-2023\)
3a-b=1/2(a+b)
=>6a-2b=a+b
=>5a=3b
=>a/3=b/5=k
=>a=3k; b=5k
\(A=\dfrac{a^{2022}+3^{2022}}{b^{2022}+5^{2022}}\)
\(=\dfrac{3^{2022}\left(k^{2022}+1\right)}{5^{2022}\left(k^{2022}+1\right)}=\left(\dfrac{3}{5}\right)^{2022}\)
\(\dfrac{2}{3}-\left|\dfrac{3}{4}\right|+\sqrt{\dfrac{25}{9}}-\left(\dfrac{2021}{2022}\right)^0=\dfrac{2}{3}-\dfrac{3}{4}+\dfrac{5}{3}-1=\dfrac{7}{12}\)
\(=\dfrac{2}{3}-\dfrac{3}{4}+\dfrac{5}{3}-1=\dfrac{7}{12}\)
A = \(\dfrac{1}{\left|x+1\right|+\left|x-2022\right|}\)
Đặt B = \(\left|x+1\right|+\left|x-2022\right|\)
\(\left|x-2022\right|\) = \(\left|2022-x\right|\) ⇒ B = \(\left|x+1\right|+\left|2022-x\right|\)
B =\(\left|x+1\right|+\left|2022-x\right|\) ≥ \(\left|x+1+2022-x\right|\) = 2023
B(min) = 2023 ⇔ (\(x+1\))(2022-\(x\)) \(\ge\) 0
Lập bảng ta có:
| \(x\) | -1 2022 |
| \(x+1\) | - 0 + | + |
| \(2022-x\) | + | + 0 - |
| (\(x+1\))(\(2022-x\)) | - 0 + 0 - |
Theo bảng trên ta có: B(min) = 2023 ⇔ -1 ≤ \(x\) ≤ 2022
A = \(\dfrac{1}{\left|x+1\right|+\left|x-2022\right|}\)
Vì A dương nên A(max) ⇔ B(min) ⇔ B = 2023
A(max) = \(\dfrac{1}{2023}\) ⇔ -1 ≤ \(x\) ≤ 2022
S = \(\left(1+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2020}\right)\)
= \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\right)-2.\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2020}\right)\)
= \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{1010}\right)\)
= \(\dfrac{1}{1011}+\dfrac{1}{1012}+...+\dfrac{1}{2021}\)