Ta có: 

\(\frac{1}{5}=\frac{1}{5}\)

\(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}

Ta có: \(S=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)

\(\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}

Ta có:

S=1/5+(1/13+1/14+1/15)+(1/61+1/62+1/63)<1/5+1/12.3+1/60.3

=>S<1/5+1/4+1/20=10/20

Hay S<1/2

Ta có: \(A=\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\)

\(A=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{62}+\frac{1}{62}+\frac{1}{63}\right)\)

\(A=\frac{1}{5}+\frac{1}{15}.3+\frac{1}{63}.3\)

\(A=\frac{1}{5}+\frac{1}{5}+\frac{1}{21}\)

\(A=\frac{47}{105}\)

Mà: \(\frac{47}{105}< \frac{47}{94}=\frac{1}{2}\)

Nên \(A=\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{2}\)

Ta có : S = 1/5 + 

cho mình xin k nha

Ta có : S = 1/5 + ( 1/13 + 1/14 + 1/15 ) + ( 1/61 + 1/62 + 1/63 ) < 1/5 + 1/12 x 3 + 1/60 x 3

S < 1/5 + 1/4 + 1/20 = 10/20 = 1/2

S < 1/2

vừa nãy ấn nhầm k mk nha

TA có:

1/12>1/13

1/12>1/14

1/12>1/15

=>1/12.3=1/4>1/13+1/14+1/15

1/60>1/61

1/60>1/62

1/60>1/63

=>1/60.3=1/20>1/61+1/62+1/63

=>1/5+1/4+1/20> 1/5+1/13+1/14+1/15+1/61+1/62+1/63

=>1/2> 1/5+1/13+1/14+1/15+1/61+1/62+1/63

bài giải:

đặt biểu thức bằng A

=> A= \(\dfrac{1}{5}+\left(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\right)\)

ta thấy:\(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< 3.\dfrac{1}{13}\)

\(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< 3.\dfrac{1}{61}\)

=> A<\(\dfrac{1}{5}+\dfrac{3}{13}+\dfrac{3}{61}\)<\(\dfrac{1}{2}\)

=> đpcm.