Question

a) Cho a\(\in\) N. Chứng minh rằng \(\dfrac{1}{a}-\dfrac{1}{a+1}< \dfrac{1}{a^2}< \dfrac{1}{a-1}-\dfrac{1}{a}\)

Answer 1

Ta có: \(\dfrac{1}{a}-\dfrac{1}{a+1}=\dfrac{a+1-a}{a\left(a+1\right)}=\dfrac{1}{a^2+a}\)

\(\dfrac{1}{a-1}-\dfrac{1}{a}=\dfrac{a-\left(a-1\right)}{\left(a-1\right)a}=\dfrac{a-a+1}{a^2-a}=\dfrac{1}{a^2-a}\)

Do \(a\in N\) nên:

\(\dfrac{1}{a^2+a}< \dfrac{1}{a^2}< \dfrac{1}{a^2-a}\)

hay \(\dfrac{1}{a}-\dfrac{1}{a+1}< \dfrac{1}{a^2}< \dfrac{1}{a-1}-\dfrac{1}{a}\left(đpcm\right)\)

Vậy...

Answer 2

Ta có:

\(\dfrac{1}{a}-\dfrac{1}{a+1}=\dfrac{a+1-a}{a\left(a+1\right)}=\dfrac{1}{a\left(a+1\right)}< \dfrac{1}{a.a}=\dfrac{1}{a^2}\left(1\right)\)

\(\dfrac{1}{a-1}-\dfrac{1}{a}=\dfrac{a-\left(a-1\right)}{a\left(a-1\right)}=\dfrac{1}{a\left(a-1\right)}>\dfrac{1}{a.a}=\dfrac{1}{a^2}\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra:

\(\dfrac{1}{a\left(a+1\right)}< \dfrac{1}{a^2}< \dfrac{1}{a\left(a-1\right)}\)

Hay \(\dfrac{1}{a}-\dfrac{1}{a+1}< \dfrac{1}{a^2}< \dfrac{1}{a-1}-\dfrac{1}{a}\) (Đpcm)