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

thế mày biết làm thì làm hộ tao cái, đéo cóp bài đứa trên

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

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

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

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

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

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

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

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

đặt A = 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 + ... + 1/100 - 2(1/2 + 1/4 + 1/6 + .... + 1/100)

= 1 + 1/2 + 1/3 + ... + 1/100 - 1 - 1/2 - 13 - ... - 1/50

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

thay vào ra E = 1

Biến đổi mẫu ta được:

\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

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

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

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

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

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

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

\(\Rightarrow P=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)\(\Rightarrow P=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

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

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

\(\Rightarrow P=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

Vậy E = 1