\(\frac{x+4}{2017}+\frac{x+3}{2018}+\frac{x+2}{2019}+\frac{x+1}{2020}\)
So sánh phân số sau:
\(\frac{2017\cdot2018}{2017\cdot2018+1}\)và\(\frac{2018\cdot2019}{2018\cdot2019+1}\)
1,So sánh A và B
\(A=\frac{2017\cdot2018-1}{2017\cdot2018}\)
\(B=\frac{2018\cdot2019-1}{2018\cdot2019}\)
tính nhanh
\(\frac{19}{37}+\left(1-\frac{19}{37}\right)\)
\(\frac{7}{13}\cdot\frac{5}{14}\cdot\frac{39}{15}\)
\(2\frac{3}{7}\cdot\frac{1}{2}-\frac{1}{2}\cdot\frac{3}{7}+\frac{1}{3}\)
\(\frac{9}{5}:\frac{17}{15}+\frac{8}{5}:\frac{17}{15}\)
\(\frac{2017}{2018}\cdot\frac{1}{2019}+\frac{2017}{2018}:\frac{2019}{2018}+\frac{1}{2018}\)
\(\frac{637\cdot527-189}{526\cdot637+448}\)
\(\frac{4}{5\cdot7}+\frac{4}{7\cdot9}+\frac{4}{9\cdot11}+...+\frac{4}{23\cdot25}\)
dấu . là dấu nhân nha mọi người
\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}\right).x+2017=\frac{2018}{1}+\frac{2018}{2}+\frac{2020}{3}\). Tìm x
Tìm x biết
a) \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)
b) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)
Cho b=\(1-\frac{2017}{2019}+\left(\frac{2017}{2019}\right)^2-\left(\frac{2017}{2019}\right)^3+...+\left(\frac{2017}{2019}\right)^{2020}\)
Cmr b<0
A=\(\frac{1+2018×2019}{2017×2019+2020}\)
\(\frac{1+2018×2019}{2017×2019+2020}\)