Question

[Cao Chí Hiếu]

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, \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+1\right):2}=1\frac{1991}{1993}\)

Answer 1

[nguyển văn hải]

A ) \(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+.....+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}.\)

=\(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)\)=101/1540

=\(\frac{101}{1540}:\frac{1}{3}=\frac{1}{5}-\frac{1}{x+3}\)

=tới đó bn tự tính nhé