Question

Cho A=\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{308}+\frac{1}{309}\)

B=\(\frac{308}{1}+\frac{307}{2}+\frac{306}{3}+...+\frac{3}{306}+\frac{2}{307}+\frac{1}{308}\). Tính \(\frac{A}{B}\)

Answer 1

A = \(\frac{1}{2}\)+ \(\frac{1}{3}\)+ \(\frac{1}{4}\)+ ... + \(\frac{1}{308}\)+ \(\frac{1}{309}\)

B = \(\frac{308}{1}\)+ \(\frac{307}{2}\)+ \(\frac{306}{3}\)+\(\frac{3}{306}\) + \(\frac{2}{307}\)+ \(\frac{1}{308}\)

=> B = \(\frac{309-1}{1}\)+ \(\frac{309-3}{3}\)+... + ( 309 ... )

=> B = 309 + 309 . ( \(\frac{1}{2}\) + \(\frac{1}{3}\)+... + \(\frac{1}{306}\)+ \(\frac{1}{307}\)+ \(\frac{1}{308}\)+ \(\frac{1}{309}\)- \(\frac{1}{1}\)+ \(\frac{2}{2}\)+ ... + \(\frac{308}{308}\)+ \(\frac{309}{309}\)

=> B = 309 . ( \(\frac{1}{2}\)+ \(\frac{1}{3}\)+ ... + \(\frac{1}{306}\)+ \(\frac{1}{307}\)+ \(\frac{1}{308}\)+ \(\frac{1}{309}\))

=> \(\frac{A}{B}\)= \(\frac{1}{309}\)

Answer 2

Lâu rồi bạn còn cần lời giải ko mình giải cho