1/97 chứ sao lại 1/91!

giải:

đặt :1/5+1/14+1/28+1/44+1/61+1/85+1/97 =A

ta có :A=1/5(1/14+1/28+1/44)+(1/61+1/85+1/97)

A<1/5(1/14.3)+(1/61.3)

A<1/5+3/14+3/61

A<1/5+3/12+1/20

A<1/5+1/4+1/20

=>A<1/2

VẬY dpcm

Cách 1: Tính hết kết quả vế trái là so sánh được => đpcm 
Cách 2: Ta đánh giá: Cho a, b là 2 số dương nếu a < b thì 1/a > 1/b 
Vậy: 
VT < 1/5 + 1/14 + 1/14 + 1/14 + 1/14 + 1/14 
= 1/5 + 5/14 = (14 + 25)/(5.14) = 39/70 < 1 (đpcm) 
Có thể còn cách khác, bạn tìm thêm đi.

Sai đề. Sửa đề :v

Cmr: \(\dfrac{1}{5}+\dfrac{1}{14}+\dfrac{1}{28}+\dfrac{1}{44}+\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{97}< \dfrac{1}{2}\)

Giải:

Đặt \(A=\dfrac{1}{5}+\dfrac{1}{14}+\dfrac{1}{28}+\dfrac{1}{44}+\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{97}\)

Ta có:

\(A=\dfrac{1}{5}+\left(\dfrac{1}{14}+\dfrac{1}{28}+\dfrac{1}{44}\right)+\left(\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{97}\right)\)

\(A< \dfrac{1}{5}\left(\dfrac{1}{14.3}\right)+\left(\dfrac{1}{61.3}\right)\)

\(A< \dfrac{1}{5}+\dfrac{3}{14}+\dfrac{3}{61}\)

\(A< \dfrac{1}{5}+\dfrac{3}{12}+\dfrac{1}{20}\)

\(A< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}\)

\(\Rightarrow A< \dfrac{1}{2}\)

Vậy \(\dfrac{1}{5}+\dfrac{1}{14}+\dfrac{1}{28}+\dfrac{1}{44}+\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{97}< \dfrac{1}{2}\) \((đpcm)\)

Ta có \(\frac{1}{5}=\frac{1}{5}\)

\(\frac{1}{14}< \frac{1}{10};\frac{1}{28}< \frac{1}{10}\)

\(\frac{1}{44}< \frac{1}{40};\frac{1}{61}< \frac{1}{40};\frac{1}{85}< \frac{1}{40};\frac{1}{97}< \frac{1}{40}\)

\(\Rightarrow\frac{1}{5}+\frac{1}{14}+\frac{1}{28}+\frac{1}{44}+\frac{1}{61}+\frac{1}{85}+\frac{1}{97}< \frac{1}{5}+\frac{1}{10}+\frac{1}{10}+\frac{1}{40}+\frac{1}{40}+\frac{1}{40}+\frac{1}{40}=\frac{1}{5}+\frac{1}{5}+\frac{1}{10}=\frac{5}{10}=\frac{1}{2}\)\(\Rightarrow A< \frac{1}{2}\)