Sửa đề: \(B=\left(\dfrac{2008}{2023}-\dfrac{2023}{2008}\right)-\left(\dfrac{-15}{2003}-\dfrac{15}{2008}\right)\)
\(=\dfrac{2008}{2023}-\dfrac{2023}{2008}+\dfrac{15}{2003}+\dfrac{15}{2008}\)
=1-1
=0
\(\dfrac{2022}{2023}.\left(\dfrac{9}{13}-\dfrac{7}{11}\right)+\dfrac{2022}{2023}.\left(\dfrac{17}{13}-\dfrac{4}{17}\right)\)
\(\left(\dfrac{1}{3}\right)^2-\left(\dfrac{1}{9}-\dfrac{2023}{2024}\right)\)
tìm x biết
\(\dfrac{x-2023}{6}\)\(+\dfrac{x-2023}{10}\)\(+\dfrac{x-2023}{15}\)\(+\dfrac{x-2023}{21}\)=\(\dfrac{8}{21}\)
12+8.(\(\dfrac{-1}{2}\))\(^3\)+\([\left(2023\right)^0:\dfrac{1}{2}]^2\)
2. Tìm x biết:
a)\(\dfrac{2}{\left(x+2\right)\left(x+4\right)}\) + \(\dfrac{4}{\left(x+4\right)\left(x+8\right)}\) + \(\dfrac{6}{\left(x+8\right)\left(x+14\right)}\) = \(\dfrac{x}{\left(x+2\right)\left(x+14\right)}\).
b)\(\dfrac{x}{2023}\) + \(\dfrac{x+1}{2022}\) + \(\dfrac{x+2}{2021}\) +...+ \(\dfrac{x+2022}{1}\) + 2023 = 0.
Gíup mình giải 2 bài này với!
Cảm ơn các bạn rất nhiều!!!
\(b,B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2022}\right)\left(1-\dfrac{1}{2023}\right)\)
help với mấy người đẹp trai xinh gái và đừng làm như anh Nguyễn Lê Phước Thịnh
tìm GTNN hoặc GTLN của D = \(\dfrac{\left|x\right|+2023}{\left|x\right|+2022}\)
\(\dfrac{x-5}{3}=\dfrac{-12}{5-x}\)
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2023}{2024}\)
Cho dãy tỉ số bằng nhau \(\dfrac{a}{2019}\) = \(\dfrac{b}{2021}\) = \(\dfrac{c}{2023}\). Chứng minh rằng \(\dfrac{\left(a-c\right)^2}{4}\) = (a - b)(b - c)
