Question

CỨU TỚ VỚI!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Chứng minh rằng:

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

AI ĐÚNG MK TK~

Answer 1

\(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{80}\)

Tách tổng trên thành 2 nhóm, ta được :

\(\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}\right)+\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}\right)\)

Mà \(\frac{1}{41}>\frac{1}{42}>...>\frac{1}{60}\); \(\frac{1}{61}>\frac{1}{62}>...>\frac{1}{80}\)

\(\Rightarrow\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}\right)+\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}\right)\)

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