đặt A=\(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{25}+\dfrac{1}{41}+\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{113}\)

        = \(\dfrac{1}{5}+(\dfrac{1}{13}+\dfrac{1}{25}+\dfrac{1}{41})+(\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{113})\)

=> A< \(\dfrac{1}{5}+(\dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12})+(\dfrac{1}{60}+\dfrac{1}{60}+\dfrac{1}{60})\)

     A<\(\dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}\)=\(\dfrac{1}{2}\)

vậy A<\(\dfrac{1}{2}\),<2=> A<2 (đpcm)

xem lại đề,1/5/1/13 là sao bạn,có phải là 1/5+1/13 không

1/5+1/13+1/25+1/41+1/61+1/85+1/113

=1/5+(1/13+1/25+1/41)+(1/85+1/61+1/113)<15+1/12+1/12+1/12+1/60+1/60+1/60

..............................................................<1/5+1/4+1/20

..............................................................<4/20+5/20+1/20

..............................................................<1/2

Cảm ơn nhiều (đúng lúc đang cần,hì ,hì)

dung luc can

\(\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+\frac{1}{41}+\frac{1}{61}+\frac{1}{85}+\frac{1}{113}=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{25}+\frac{1}{41}\right)+\left(\frac{1}{61}+\frac{1}{85}+\frac{1}{113}\right)\)

 < \(\frac{1}{5}+\frac{1}{12}.3+\frac{1}{60}.3=\frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{4}{20}+\frac{5}{20}+\frac{1}{20}=\frac{10}{20}=\frac{1}{2}\)(đpcm)

ê cho hỏi tại sao lại ra < \(\frac{1}{5}+\frac{1}{12}.3+\frac{1}{60}.3\)

VÌ BÉ HƠN MÀ BN

a, 1/5+1/6+1/7+1/8+1/9 < 1/5.5=1  ⁽¹⁾

    1/10+1/11+1/12+1/13+1/14+1/15+1/16+1/17 < 1/10.7 < 1/10.10 < 1  ⁽²⁾

    Từ (1) và (2) , suy ra 1/5+1/6+1/7+...+1/17 < 1+1 =2

Suy ra , 1/5+1/6+1/7+...+1/17 < 2

b, Ta cần c/m 1/13+1/25+1/41+1/61+1/85+1/113 < 3/10 (Vì 1/2 - 1/5 = 3/10)

                     1/13+1/25+1/41+1/61+1/85+1/113 < 1/10+1/25+1/25+1/25+1/25+1/25

                     1/13+1/25+1/41+1/61+1/85+1/113 < 1/10 + 5/25 = 1/10+1/5 = 3/10

Suy ra , 1/5+1/13+1/25+1/41+1/61+1/85+1/113 < 1/2