\(\left|-1\dfrac{1}{2}+1,2\right|+1\dfrac{2}{3}:6-1^{2022}\)
\(=\dfrac{7}{10}+\dfrac{5}{3}:6-1\)
\(=\dfrac{7}{10}+\dfrac{5}{18}-1\)
\(=\dfrac{44}{45}-1\)
\(=\dfrac{-1}{45}\)
\(\left|-1\dfrac{1}{2}+1,2\right|+1\dfrac{2}{3}:6-1^{2022}\)
\(=\dfrac{7}{10}+\dfrac{5}{3}:6-1\)
\(=\dfrac{7}{10}+\dfrac{5}{18}-1\)
\(=\dfrac{44}{45}-1\)
\(=\dfrac{-1}{45}\)
Cho 2022 số tự nhiên a(1), a(2), a(3), ..., a(2021), a(2022) khác 0 thỏa mãn:
\(\dfrac{1}{a\left(1\right)}\) + \(\dfrac{1}{a\left(2\right)}\) + ... + \(\dfrac{1}{a\left(2021\right)}\) + \(\dfrac{1}{a\left(2022\right)}\) = 1. Chứng minh rằng: tồn tại ít nhất một số trong 2022 số đã cho là số chẵn.
Tính M=\(\left(\dfrac{0,4-\dfrac{2}{9}+\dfrac{2}{11}}{1,4-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-0,25+\dfrac{1}{5}}{1\dfrac{1}{6}-0,875+0,7}\right):\dfrac{2021}{2022}\)
\(tínhhợplí:\left(1,2-\sqrt{\dfrac{1}{4}:1\dfrac{1}{20}+\left|-\dfrac{3}{4}\right|-\left(-\dfrac{3}{2}\right)^2}\right)\)
Bài 1:
\(S=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\) và \(P=\dfrac{1}{1012}+\dfrac{1}{1013}+...+\dfrac{1}{2022}\)
Tính \(\left(S-P\right)^{2022}\)
Mọi người giúp mình với, mình cảm ơn !!!
a, \(\left(2x-1\right)\left(x+\dfrac{2}{3}\right)=0\)
b, \(\dfrac{x+4}{2019}+\dfrac{x+3}{2020}=\dfrac{x+2}{2021}+\dfrac{x+1}{2022}\)
\(b,B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2022}\right)\left(1-\dfrac{1}{2023}\right)\)
help với mấy người đẹp trai xinh gái và đừng làm như anh Nguyễn Lê Phước Thịnh
1. a) \(\dfrac{5}{3}+\dfrac{-2}{7}-\left(-1,2\right)\)
b) \(0,25+\dfrac{3}{5}-\left(\dfrac{1}{8}-\dfrac{2}{5}+1\dfrac{1}{4}\right)\)
Giúp mình với ![]()
Thực hiện phép tính:
a) 2021 - \(\left(\dfrac{1}{3}\right)^2\) . 32
b \(\dfrac{5}{10}\) + 9 . \(\dfrac{-3}{2}\)
c) -10 . \(\left(-\dfrac{2021}{2022}\right)^0\) + \(\left(\dfrac{2}{5}\right)^2\) : 2
(\(\sqrt{\dfrac{1}{4}}\) - 1,2) : 1\(\dfrac{1}{20}\) - (- \(\dfrac{5}{2}\))2 + \(\left|1,25-\dfrac{3}{4}\right|\)
Tính : S = \(1-\dfrac{1}{2}+\dfrac{1}{3}-\)\(\dfrac{1}{4}+...+\dfrac{1}{2019}-\dfrac{1}{2020}+\dfrac{1}{2021}\)và
P = \(\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}+...+\dfrac{1}{2020}+\dfrac{1}{2021}\)
Tính : \(\left(S-P\right)^{2022}\)