A = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

=\(1-\frac{1}{50}\)

Vì \(1-\frac{1}{50}< 1\)nên A < 1

B = \(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

=\(\frac{1}{2}-\frac{1}{100}\)

Vì \(\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)nên B < \(\frac{1}{2}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=1-\frac{1}{50}\)

\(\Rightarrow A< 1\)

\(B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

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

\(B=\frac{1}{2}-\frac{1}{100}\)

\(\Rightarrow B< \frac{1}{2}\)

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=1-\frac{1}{50}\)

\(A=\frac{49}{50}< 1\)

\(B=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\)

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

\(B=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)

ta có:

 \(\frac{1}{11}\)>\(\frac{10}{20}\)

\(\frac{1}{12}\)>\(\frac{10}{20}\)

\(\frac{1}{13}\)>\(\frac{10}{20}\)

....

\(\frac{1}{19}\)>\(\frac{10}{20}\)

=>E >\(\frac{10}{20}\)

vậy E > \(\frac{1}{2}\)

trên violympic phải ko, mình vừa mới giải xong nè

\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.... +\frac{1}{100}-\frac{1}{101}\)

Ta có:

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\)

\(\Rightarrow A=1-\frac{1}{2}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\)

\(\Rightarrow A=1-\frac{1}{101}=\frac{100}{101}< 1\)

Vậy : \(A< 1\)

~ Rất vui vì giúp đc bn ~ ^_<

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{100.101}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{100}-\frac{1}{101}\)

\(A=\left(1-\frac{1}{101}\right)+\left(-\frac{1}{2}+\frac{1}{2}\right)+...+\left(-\frac{1}{100}+\frac{1}{100}\right)\)

\(A=1-\frac{1}{101}\)

\(A=\frac{100}{101}< 1\)

=> A <1

Ta có

\(M=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^3.3^2}+.....+\frac{100^2-99^2}{99^2.100^2}\)

\(M=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+......+\frac{1}{99^2}-\frac{1}{100^2}\)

\(M=1-\frac{1}{100^2}< 1\)

=> M<1

\(M=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{100^2-99^2}{99^2.100^2}\)

\(M=\frac{2^2}{1^2.2^2}-\frac{1^2}{1^2.2^2}+\frac{3^2}{2^2.3^2}-\frac{2^2}{2^2.3^2}+\frac{4^2}{3^2.4^2}-\frac{3^2}{3^2.4^2}+...+\frac{100^2}{99^2.100^2}-\frac{99^2}{99^2.100^2}\)

\(M=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{99^2}-\frac{1}{100^2}=1-\frac{1}{100^2}

\(S=\frac{3}{1^2\cdot2^2}+\frac{5}{2^2\cdot3^2}+\frac{7}{3^2\cdot4^2}+...+\frac{99}{49^2\cdot50^2}\)

\(=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+.....+\frac{1}{49^2}-\frac{1}{50^2}\)

\(=1-\frac{1}{50^2}=\frac{2499}{2500}\)

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

\(=\frac{1}{1\cdot3}+\frac{1}{2\cdot4}+\frac{1}{3\cdot5}+...+\frac{1}{49\cdot51}\)

\(=\frac{1}{2}\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(=\frac{1}{2}\cdot\left(1+\frac{1}{2}-\frac{1}{51}\right)=\frac{151}{204}\)

Vì \(\frac{2499}{2500}>\frac{151}{204}\)nên S>T

JOKER_Võ Văn Quốc, T = \(\frac{1}{2}.\left(1-\frac{1}{51}+\frac{1}{2}-\frac{1}{50}\right)\)mới đúng
Sẽ dễ hơn nếu bạn chia ra 2 vế \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)và \(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{48+50}\)

M\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{49}-\frac{1}{50}\)

\(1-\frac{1}{50}=\frac{49}{50}\)

vì \(\frac{49}{50}

=>M=1/1-1/2+1/2-1/3+....+1/49-1/50
=>M=1/1-1/50
=>M=0.98

B = \(\frac{1}{2^2.3-5^2.7}\)= \(\frac{-1}{163}\)

Đặt A = \(\frac{1}{101^2}+\frac{1}{102^2}+...+\frac{1}{105^2}\)

Vì A > 0 (các số hạng của A đều > 0)

Mà B < 0

=> A > B

-Ta có:A= 1/101^2+1/102^2+1/103^2+1/104^2+1/105^2 
         A>1/(100x101)+1/(101x102)+1/(102x103)+... 

-Vì cùng tử mẫu nhỏ hơn thì lớn hơn
         A>1/100-1/101+1/101-1/102+1/102-1/103+... 
         A>1/100-1/105=1/2100=1/(2^2.3.5^2.7)=B 
=>Vậy A>B