=> A < B

chắc vại-.-

tui hok giỏi toán lém

con bái mẹ 

Answer:

Mình làm thành tính tỉ số luôn nhé!

\(A=\frac{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}{\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}}\)

Ta xét \(B=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

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

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

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

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

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

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

\(=\left(1-1\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+...+\left(\frac{1}{50}-\frac{1}{50}\right)+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

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

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

\(\Rightarrow\frac{A}{B}=1\)

2.2=4. đúng nên tick nha!

Xét mẫu số:   1/(2x3) + 1/(3x4) + …… + 1/(99x100)

       = 1/1 – 1/2 + 1/3 – 1/4 + ......... + 1/99 – 1/100

       = (1 + 1/3 + ............ + 1/99) – (1/2 + 1/4 + .......... + 1/100)

       = (1 + 1/3 + ............ + 1/99)+(1/2+1/4+1/6+….+1/100) – (1/2+1/4+1/6+ .......... + 1/100)x2

       = (1 + 1/2 + 1/3 + 1/4 + ..... + 1/99 + 1/100) – (1 + 1/2  + 1/3 + ....... +1/50 )

       = 1/51 + 1/52 + 1/53 + ............. + 1/100            (Đơn giản số trừ)

Vậy:  (1/51 + 1/52 + 1/53 + ............. + 1/100) / (1/1x2 + 1/3x4 + .......... + 1/99x100)     =

          (1/51 + 1/52 + 1/53 + ............. + 1/100) / (1/51 + 1/52 + 1/53 + ............. + 1/100) = 1

Mày ko duyệt thì CKET

B=1/1.2+1/3.4+1/5.6+...+1/99.100

=1-1/2+1/3-1/4+1/5-1/6+...+1/99-1/100

=(1+1/3+1/5+...+1/99)-(1/2+1/4+1/6+...+1/100)

=(1+1/2+1/3+1/4+1/5+1/6+...+1/99+1/100)-2(1/2+1/4+1/6+...+1/100)

=(1+1/2+1/3+1/4+...+1/100)-(1+1/2+1/3+..+1/50)

=1/51+1/52+1/53+..+1/100 (1)

A=1/51+1/52+1/53+..+1/100 (2)

(1),(2)=> A/B=1

\(\frac{A}{B}=1\)

\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{99\cdot100}\)

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

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

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

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

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

Ta có : \(VT=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+...+\frac{1}{99.100}\)

               \(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{99}-\frac{1}{100}\)

                \(=\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{100}\right)\)

                \(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+...+\frac{1}{100}\right)\) 

                 \(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{50}\right)\)

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

\(\Rightarrow\) \(ĐPCM\)

mik nhớ kq là ..........50 thì phải

45854

212122512122

1

1

1

1123

4564

454

3546434

\(C=\frac{1}{1.2}+\frac{1}{3.4}+.....+\frac{1}{99.100}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{99}-\frac{1}{100}=\left(1+\frac{1}{3}+....+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{100}\right)=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{100}\right)=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}....+\frac{1}{50}\right)=\frac{1}{51}+\frac{1}{52}+....+\frac{1}{100}\Rightarrow C:D=1\)