\(\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: 

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

\(\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)

Giải:

Ta có:

\(S=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\)

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

Nhận xét:

\(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< \dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}=\dfrac{1}{4}\)

\(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{60}+\dfrac{1}{60}+\dfrac{1}{60}=\dfrac{1}{20}\)

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

Vậy \(S=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\) \(< \dfrac{1}{2}\) (Đpcm)

\(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{5}+\dfrac{1}{13}\cdot3+\dfrac{1}{61}\cdot3\\ =\dfrac{1}{5}+\dfrac{3}{13}+\dfrac{3}{61}< \dfrac{1}{5}+\dfrac{3}{12}+\dfrac{3}{60}=\dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}=\dfrac{1}{2}\)

=> Điều phải chứng minh

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 : \(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)

Ta có:

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

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

Bài toán phụ 1:

Ta có:

1/13<1/12

1/14<1/12

1/15<1/12

=>1/13+1/14+1/15<1/12x3=1/4 (1)

Bài toán phụ 2:

Ta có:

1/61<1/60

1/62<1/60

1/63<1/60

=>1/61+1/62+1/63<1/60x3=1/20 (2)

Từ (1) và (2), ta có:

1/5+1/13+1/14+1/15+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<4/20+5/20+1/20

1/5+1/13+1/14+1/15+1/61+1/62+1/63<9/20<1/2

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

Ta có :

\(S=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\)

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

Nhận xét :

\(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< \dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}=\dfrac{1}{4}\)

\(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{60}+\dfrac{1}{60}+\dfrac{1}{60}=\dfrac{1}{20}\)

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

\(\Rightarrow S< \dfrac{1}{2}\rightarrowđpcm\)