Question

Cho \(b\in N\)*, b < 1. Chứng minh \(\dfrac{1}{b}-\dfrac{1}{b+1}< \dfrac{1}{b^2}< \dfrac{1}{b-1}-\dfrac{1}{b}\)

Answer 1

tìm trước khi hỏi

Answer 2

Xét \(\dfrac{1}{b}-\dfrac{1}{b+1}=\dfrac{b+1-b}{b\left(b+1\right)}=\dfrac{1}{b\left(b+1\right)}\)

vì b+1 > b , b \(\in\) N sao , => b(b+1) >b^{2 => \(\dfrac{1}{b}-\dfrac{1}{b+1}< \dfrac{1}{b^2}\)}

Xét \(\dfrac{1}{b-1}-\dfrac{1}{b}=\dfrac{b-b+1}{b\left(b-1\right)}=\dfrac{1}{b\left(b-1\right)}\)vì b>b-1

=> b²>(b-1)b => \(\dfrac{1}{b^2}< \dfrac{1}{b\left(b-1\right)}\)

Vậy\(\dfrac{1}{b}-\dfrac{1}{b+1}< \dfrac{1}{b^2}< \dfrac{1}{b-1}-\dfrac{1}{b}\left(đpcm\right)\)