Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Ta có:
\(x-\left\{\left[-x-\left(x+3\right)\right]-\left[\left(x+2018\right)-\left(x+2019\right)\right]+21\right\}\)
\(=x-\left\{\left[-x-x-3\right]-\left[x+2018-x-2019\right]+21\right\}\)
\(=x-\left\{\left[-2x-3\right]-\left[2018-2019\right]+21\right\}\)
\(=x+2x+-3+1-21\)
\(=3x-23\)
=> \(3x-23=2020\)
\(3x=2020+23=2043\)
=> \(x=2043:3=681\)
Nhầm
\(=x-\left\{-2x-3+1+21\right\}\\ =x+2x+3-1-21\)
\(=3x-17\\ =>3x-17=2020\\ 3x=2020+17=2037\\ x=2037:3=679\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x(x+1)}=\frac{2019}{2020}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{2020}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{2019}{2020}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{2019}{2020}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2020}\)
\(\Rightarrow x+1=2020\Leftrightarrow x=2019\)
Vậy x = 2019
Bạn xem lại. Biểu thức C có GTLN chứ không có GTNN bạn nhé.
Ta có: \(A=\left(2020^{2019}+2019^{2019}\right)^{2020}\)
\(=\left(2019^{2019}+2020^{2019}\right)^{2019}\cdot\left(2019^{2019}+2020^{2019}\right)\)
\(\Leftrightarrow\dfrac{A}{B}=\dfrac{\left(2019^{2019}+2020^{2019}\right)^{2019}\cdot\left(2019^{2019}+2020^{2019}\right)}{\left(2020^{2020}+2019^{2020}\right)^{2019}}\)
\(\Leftrightarrow\dfrac{A}{B}=\dfrac{2019^{2019}+2020^{2019}}{2019+2020}>1\)
\(\Leftrightarrow A>B\)
