\(A=1-\frac{2}{3}+1-\frac{2}{15}+1-\frac{2}{35}+1-\frac{2}{63}+1-\frac{2}{99}+1-\frac{2}{143}\)      

\(=1+1+1+1+1+1-\frac{2}{3}-\frac{2}{15}-\frac{2}{35}-\frac{2}{63}-\frac{2}{99}-\frac{2}{143}\)   

\(=6-\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\right)\)   

\(=6-\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)\)   

\(=6-\left(1-\frac{1}{13}\right)\)   

\(=6-1+\frac{1}{13}\)   

\(=5+\frac{1}{13}\)   

\(=\frac{65}{13}+\frac{1}{13}\)   

\(=\frac{66}{13}\)

\(A=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\)

\(A=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}\)

\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\)

\(A=\frac{1}{3}-\frac{1}{13}\)

\(A=\frac{10}{39}\)

      \(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\)

\(=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\)

\(=\frac{1}{3}-\frac{1}{13}\)

\(=\frac{10}{39}\)

_Hok tốt_

!!!

A = \(\frac{1}{3}+\frac{13}{35}+\frac{33}{35}+\frac{61}{63}+\frac{97}{99}+\frac{141}{143}\)

\(=\left(1-\frac{2}{3}\right)+\left(1-\frac{2}{15}\right)+\left(1-\frac{2}{35}\right)+\left(1-\frac{2}{63}\right)+\left(1-\frac{2}{99}\right)+\left(1-\frac{2}{143}\right)\)

\(=\left(1+1+1+1+1+1\right)-\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\right)\)

\(=6-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)

\(=6-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)

\(=6-\left(1-\frac{1}{13}\right)\)

\(=6-1+\frac{1}{13}\)

\(=5+\frac{1}{13}\)

\(=\frac{66}{13}\)

\(\text{Vậy }A=\frac{66}{13}\)

dễ lm bạn ơi

B=4/3.5+4/5.7+.........................

Ghi tới đok chắc hỉu roài chứ, tự lm phần sau nhé.

Cho A chứng minh B, kì vậy ?

Ta có: \(A=\frac{3-2}{3}+\frac{15-2}{15}+\frac{35-2}{35}+\frac{63-2}{63}+\frac{99-2}{99}+\frac{143-2}{143}+\frac{195-2}{195}\)

\(A=\left(1-\frac{2}{3}\right)+\left(1-\frac{2}{15}\right)+\left(1-\frac{2}{35}\right)+\left(1-\frac{2}{63}\right)+\left(1-\frac{2}{99}\right)+\left(1-\frac{2}{143}\right)+\left(1-\frac{2}{195}\right)\)

\(A=7-\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}+\frac{2}{195}\right)\)

\(A=7-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\right)\)

\(A=7-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)

\(A=7-\left(1-\frac{1}{15}\right)=7-1+\frac{1}{15}=6\frac{1}{15}\)không là số nguyên

a,

\(A=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{900}\right)\\ =\left(1-\frac{1}{2}\right)\left(1+\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1+\frac{1}{3}\right)...\left(1-\frac{1}{30}\right)\left(1+\frac{1}{30}\right)\\ =\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot...\cdot\frac{29}{30}\cdot\frac{31}{30}\\ =\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{29}{30}\cdot\frac{3}{2}\cdot\frac{4}{3}\cdot...\cdot\frac{31}{30}\\ =\frac{1\cdot2\cdot...\cdot29}{2\cdot3\cdot...\cdot30}\cdot\frac{3\cdot4\cdot...\cdot31}{2\cdot3\cdot...\cdot30}\\ =\frac{1}{30}\cdot\frac{31}{2}=\frac{31}{60}\)

b,

\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{98\cdot99\cdot100}\right)\\ =\frac{1}{2}\left(\frac{3-1}{1\cdot2\cdot3}+\frac{4-2}{2\cdot3\cdot4}+...+\frac{100-98}{98\cdot99\cdot100}\right)\\ =\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\\ =\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)\\ =\frac{1}{2}\cdot\frac{4450-1}{9900}=\frac{1}{2}\cdot\frac{4449}{9900}=\frac{4449}{19800}=\frac{1483}{6600}\)

c, (Chịu :V)

d,

\(D=\frac{1}{3}\left(\frac{3}{1\cdot2\cdot3\cdot4}+\frac{3}{2\cdot3\cdot4\cdot5}+...+\frac{3}{27\cdot28\cdot29\cdot30}\right)\\ =\frac{1}{3}\left(\frac{4-1}{1\cdot2\cdot3\cdot4}+\frac{5-2}{2\cdot3\cdot4\cdot5}+...+\frac{30-27}{27\cdot28\cdot29\cdot30}\right)\\ =\frac{1}{3}\left(\frac{1}{1\cdot2\cdot3}-\frac{1}{2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4}-\frac{1}{3\cdot4\cdot5}+...+\frac{1}{27\cdot28\cdot29}-\frac{1}{28\cdot29\cdot30}\right)\\ =\frac{1}{3}\left(\frac{1}{6}-\frac{1}{24630}\right)\\ =\frac{228}{4105}\)

Chúc bạn học tốt nha.

a) Đề thiếu nhé. sửa đề:

 \(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{195}\)

\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\)

\(=1-\frac{1}{15}\)

\(=\frac{14}{15}\)