Question

d, \(\frac{1}{1.5}+\frac{1}{5.9}+...........+\frac{1}{397.401}< \frac{1}{4}\)

Answer 1

Gọi A=1/1.5+1/5.9+...+1/397.401

Ta có:

\(4A=\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{397.401}=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{397}-\frac{1}{401}=1-\frac{1}{401}\)

=>\(A=\frac{1}{4}-\frac{1}{1604}< \frac{1}{4}\)

=>đpcm

Answer 2

d, Đặt biểu thức trên là S, ta có:

S = 1/1.5 + 1/5.9 + .... + 1/397.401 < 1/4

Nhân cả hai vế với 4 ,ta có :

4S = 4. ( 1/1.5 + 1/5.9 + .... + 1/397.401 )

4S = 4/1.5 + 4/5.9 + .... + 4/397.401

4S = 1 - 1/5 + 1/5 - 1/9 + .... +1/397 - 1/401

4S = 1 - 1/401

4S = 400/401

  S = 400/401 : 4 

  S = 100/401.

Ta có : 100/401 và 1/4

            400/1604 < 401/1604

=>  S < 1/4

Vậy 1/1.5 + 1/5.9 + .... + 1/397.401 < 1/4

Answer 3

Đặt \(A=\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{397.401}\)

\(\Rightarrow4A=\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{397.401}\)

\(4A=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{397}-\frac{1}{401}\)

\(4A=1-\frac{1}{401}\)

\(A=\frac{1}{4}-\frac{1}{1604}< \frac{1}{4}\left(đpcm\right)\)

hok tốt .