Question

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

Answer 1

Sửa đề là chứng minh nha bạn.

Ta có: \(\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{80}>\dfrac{1}{41}+\dfrac{1}{41}+\dfrac{1}{41}+...+\dfrac{1}{41}\)(40 phân số \(\dfrac{1}{41}\))

\(=\dfrac{1.40}{41}=\dfrac{40}{41}>\dfrac{7}{12}\) (*)

Từ (*) suy ra: \(\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{80}>\dfrac{7}{12}^{\left(đpcm\right)}\)

Answer 2

đpcm là gì