Biết \(\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+...+\frac{1}{x\times\left(x+3\right)}=\frac{101}{1540}\) .Khi đó x=.............
Biết \(\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+...+\frac{1}{x\left(x+5\right)}=\frac{101}{1540}\)
Tìm x
Tìm x, biết rằng:
a)\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
b)\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=1\frac{1991}{1993}\)
Tìm x biết
a) \(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
b) \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=1\frac{2017}{2019}\)
Tìm x biết \(\frac{2}{40}+\frac{2}{88}+\frac{2}{154}+....+\frac{2}{x\left(x+3\right)}=\frac{202}{1540}\)
Tìm x biết:
a, \(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
b, \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)}=\frac{2021}{2023}\)
Giải nhanh giùm mình nhé!!
Tìm x biết:
\(\frac{2}{40}+\frac{2}{88}+\frac{2}{154}+...+\frac{2}{x\left(x+3\right)}=\frac{202}{1540}\)
tìm x biết :
\(\frac{2}{40}+\frac{2}{88}+\frac{2}{154}+.....+\frac{2}{x\left(x+3\right)}=\frac{202}{1540}\)
Tìm x, biết :
a, \(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x.\left(x+3\right)}=\frac{101}{1540}\)
b, \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+1\right):2}=1\frac{1991}{1993}\)