Ta có : 

\(S=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=50.\frac{1}{100}=\frac{50}{100}=\frac{1}{2}\)

\(\Rightarrow\)\(S>\frac{1}{2}\)

Vậy \(S>\frac{1}{2}\)

Chúc bạn học tốt ~

\(S>\frac{1}{100}\cdot50=\frac{1}{2}\)

vì\(\frac{1}{51}>\frac{1}{52}>....>\frac{1}{100}\)

=>\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)> 50x1/100=1/2(dpcm)

ta có 1/51>1/100

        1/52>1/100

        ..................

        1/100=1/100

\(\Rightarrow\)S=1/51+1/52+...+1/100>(1/100+1/100+...+1/100)=1/100.50=1/2

\(\Rightarrow\)S>\(\frac{1}{2}\)

cái chỗ 1/100+1/100+...+1/100 có 50 số bạn nhá

chúc bạn học tốt~

\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)

1/51+1/52+1/53 +...+1/100

S = \(\frac{1}{50}+\frac{1}{51}+...+\frac{1}{99}>\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{1}{100}.50=\frac{1}{2}\)

Kết luận vậy S > 1/2

S=1/2

S>\(\frac{1}{2}\)

Giải:

\(S=\dfrac{1}{50}+\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{98}+\dfrac{1}{99}\) 

\(S=\left(\dfrac{1}{50}+\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{74}\right)+\left(\dfrac{1}{75}+...+\dfrac{1}{98}+\dfrac{1}{99}\right)\) 

\(\Rightarrow S>\left(\dfrac{1}{50}+\dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}\right)+\left(\dfrac{1}{75}+...+\dfrac{1}{75}+\dfrac{1}{75}\right)\) 

\(\Rightarrow S>\dfrac{1}{2}+\dfrac{1}{3}>\dfrac{1}{2}\) 

\(\Rightarrow S>\dfrac{1}{2}\left(đpcm\right)\) 

Ta có:S=1/50+1/51+1/52+...+1/99

S>1/50+1/50+1/50+....+1/50(50 số hạng)

S>1/50x50

S>1>1/2

=>S>1/2

\(M=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}\)

\(\Rightarrow M< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}\)

\(\Rightarrow M< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}\)

\(\Rightarrow M< 1-\frac{1}{99}< 1\)

Dễ thấy M > 0 nên 0 < M < 1

Vậy M không là số tự nhiên.

\(S=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)

\(\Rightarrow S>\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\) (50 số hạng \(\frac{1}{100}\))

\(\Rightarrow S>\frac{1}{100}.50=\frac{1}{2}\)

Vậy \(S>\frac{1}{2}\left(đpcm\right)\)

\(S=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)

\(\Rightarrow S< \frac{1}{50}+\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\)(50 số hạng \(\frac{1}{50}\))

\(\Rightarrow S< \frac{1}{50}.50=1\)

Vậy S < 1 (đpcm)

Mình không chắc đã đúng đâu nhưng mình cứ giair thử nhé ! 

Ta có : 

A = \(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\)+ ... + \(\frac{1}{99}-\frac{1}{100}\)

= \(\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...\frac{1}{99}\right)\)- \(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}...+\frac{1}{100}\right)\)

= \(\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...\frac{1}{99}\right)\)+ \(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}...+\frac{1}{100}\right)\)

- \(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)x 2 

= \(\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)- \(\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

= \(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)= B 

Vậy , A = B 

~ Chúc bạn học giỏi ! ~

OK chơi luôn nhưng có phải cứ cách 2 phân số lai có 1 phân số có mẫu bằng 8 hay chỉ có 1 thui

7X-8=713