Question

Hãy chứng tỏ rằng :

  \(\frac{1}{41}\)+ \(\frac{1}{42}\)+ \(\frac{1}{43}\)+ ..... + \(\frac{1}{79}\)+ \(\frac{1}{80}\)> \(\frac{7}{12}\)

Các bn làm giúp mk vs, bài này khó, nghĩ mãi ko ra

Answer 1

Ta có : 

\(\frac{7}{12}\)= \(\frac{4}{12}\)+ \(\frac{3}{12}\)= \(\frac{1}{3}\)+ \(\frac{1}{4}\)= \(\frac{20}{60}\)+ \(\frac{20}{80}\)

\(\frac{1}{41}\)+ \(\frac{1}{42}\)+ \(\frac{1}{43}\)+ .... + \(\frac{1}{79}\)+ \(\frac{1}{80}\)= (\(\frac{1}{41}\)+ \(\frac{1}{42}\)+ \(\frac{1}{43}\)+ ....+\(\frac{1}{60}\)) + ( \(\frac{1}{61}\)+ \(\frac{1}{62}\)+...+\(\frac{1}{79}\)+\(\frac{1}{80}\))

Do \(\frac{1}{41}\)>\(\frac{1}{42}\)>....>\(\frac{1}{60}\)

=> ( \(\frac{1}{41}\)+ \(\frac{1}{42}\)+...+\(\frac{1}{60}\)) > \(\frac{1}{60}\)+...+\(\frac{1}{60}\)= \(\frac{20}{60}\)

Vậy : \(\frac{1}{61}\)> \(\frac{1}{62}\)>....>\(\frac{1}{79}\)>\(\frac{1}{80}\)

=> ( \(\frac{1}{61}\)+\(\frac{1}{62}\)+...+\(\frac{1}{79}\)+ \(\frac{1}{80}\)) > \(\frac{1}{80}\)+...+ \(\frac{1}{80}\)= \(\frac{20}{80}\)

Vậy : \(\frac{1}{41}\)+ \(\frac{1}{42}\)+....+\(\frac{1}{79}\)+ \(\frac{1}{80}\)> \(\frac{20}{60}\)+ \(\frac{20}{80}\)

Vậy : \(\frac{1}{41}\)+ \(\frac{1}{42}\)+....+ \(\frac{1}{79}\)+ \(\frac{1}{80}\)> \(\frac{20}{60}\)+ \(\frac{20}{80}\)= \(\frac{7}{12}\)

=> ĐPCM