S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)

Mà : (1/31+1/32+1/33+...+1/40) > 1/40 x 10 = 1/4 (gồm 10 số hạng)

Tương tự ta có : (1/41 + 1/42 + ...+ 1/50) > 1/5 ;   (1/51 + 1/52+...+1/59+1/60) > 1/6

S > 1/4 + 1/5 + 1/6.

Mà khi đó ta thấy: (1/4 + 1/5 + 1/6) > 3/5

=>S > 3/5                             (1)

S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)

Do : (1/31+1/32+1/33+...+1/40) < 1/31 x 10 = 10/30 = 1/3 (gồm 10 số hạng)

=> S <  4/5                             (2)

Từ (1) và (2) => 3/5 <S<4/5

A=1/31+1/32+...+1/149+1/150

1/31<1/30

1/32<1/30

...

1/40<1/30

1/41<1/40

1/42<1/40

...

1/50<1/40

...

1/140<1/130

1/141<1/140

...

1/150<1/140

=>A<10(1/30+1/40+...+1/140)

=>A<1/3+1/4+...+1/14=1,75<13/6

Giải:

S=\(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{60}\) 

Có 30 phân số; chia làm 3 nhóm

S<\(\left(\dfrac{1}{30}+...+\dfrac{1}{30}\right)+\left(\dfrac{1}{40}+...+\dfrac{1}{40}\right)+\left(\dfrac{1}{50}+...+\dfrac{1}{50}\right)\) 

S<\(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}\) 

S<\(\dfrac{47}{60}< \dfrac{48}{60}=\dfrac{4}{5}\) 

⇒S<\(\dfrac{4}{5}\) (đpcm)

Chúc bạn học tốt!

\(1,Y=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{96}+3^{97}+3^{98}\right)\\ Y=\left(1+3+3^2\right)\left(1+3^3+...+3^{96}\right)\\ Y=13\left(1+3^3+...+3^{96}\right)⋮13\\ 2,A=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{2018}+3^{2019}\right)\\ A=\left(1+3\right)\left(1+3^2+...+3^{2019}\right)\\ A=4\left(1+3^2+...+3^{2019}\right)⋮4\\ 3,\Leftrightarrow2\left(x+4\right)=60\Leftrightarrow x+4=30\Leftrightarrow x=36\)