Question

BT1: Chứng tỏ rằng: \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}>\dfrac{5}{6}\)

BT2: Điền vào tổng sau số còn thiếu sau đó tính tổng:

\(\dfrac{1}{5}+\dfrac{1}{45}+\dfrac{1}{117}+...+\dfrac{1}{1517}\)

Answer 1

BT1: \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}>\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}>1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{5}{6}\)

Vậy ta suy ra đpcm

Answer 2

1. Ta có :

\(\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{6}>\dfrac{1}{6}+\dfrac{1}{6}+.....+\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{6}< \dfrac{1}{6}.5\)

\(\Leftrightarrow\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{6}< \dfrac{5}{6}\)

\(\rightarrowđpcm\)