\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)

\(=1-\frac{1}{99}\)

\(=\frac{98}{99}\)

\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)

\(=1-\frac{1}{99}\)

\(=\frac{98}{99}\)

Dễ 

\(\frac{2327}{4851}\)

co  can cách  làm ko bạn      

có,bạn gửi luôn cho mình

\(2S=\frac{2}{1}-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+...+\frac{2}{97}-\frac{2}{99}\)

\(2S=2-\frac{2}{99}\)

\(2S=\frac{196}{99}\)

\(S=\frac{196}{99}\cdot\frac{1}{2}=\frac{98}{99}\)

Ta có: S=2/1.3+2/3.5+...+2/97.99

S= 2/2.(1-1/3+1/3-1/5+...+1/97-1/99)

S= 1-1/99=98/99

\(S=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)

\(2S=2\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\right)\)

\(\Rightarrow S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)

\(S=1-\frac{1}{99}\)

\(S=\frac{98}{99}\)

Đặt \(A=\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{97.99}+\frac{1}{99.100}\)

\(\Rightarrow2A=2\left(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{97.99}+\frac{1}{98.100}\right)\)

\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{2.4}+\frac{2}{3.5}+...+\frac{2}{97.99}+\frac{2}{98.100}\)

\(\Rightarrow2A=\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)+\left(\frac{2}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)

\(\Rightarrow2A=\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)+\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)

\(\Rightarrow2A=\left(1-\frac{1}{99}\right)+\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(\Rightarrow2A=\left(\frac{99}{99}-\frac{1}{99}\right)+\left(\frac{50}{100}-\frac{1}{100}\right)\)

\(\Rightarrow2A=\frac{98}{99}+\frac{49}{100}=\frac{9800}{9900}+\frac{4851}{9900}=\frac{14651}{9900}\)

\(\Rightarrow A=\frac{14651}{9900}:2=\frac{14651}{9900}.\frac{1}{2}=\frac{14651}{19800}\)

bạn nhớ thử lại nhé :)

\(\left(\frac{3}{1.3}+\frac{3}{3.5}+.......+\frac{3}{97.99}\right).\left(2x+1\right)=x+\frac{1}{33}\)

\(\Rightarrow[\frac{3}{2}.(\frac{2}{1.3}+\frac{2}{3.5}+.......+\frac{2}{97.99})].\left(2x+1\right)=x+\frac{1}{33}\)

\(\Rightarrow[\frac{3}{2}.(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{97}-\frac{1}{99})].\left(2x+1\right)=x+\frac{1}{33}\)

\(\Rightarrow[\frac{3}{2}.(1-\frac{1}{99})].\left(2x+1\right)=x+\frac{1}{33}\)

\(\Rightarrow\left(\frac{3}{2}.\frac{98}{99}\right).\left(2x+1\right)=x+\frac{1}{33}\)

\(\Rightarrow\frac{49}{33}.\left(2x+1\right)=x+\frac{1}{33}\)

\(\Rightarrow\frac{49}{33}.2x+\frac{49}{33}=x+\frac{1}{33}\)

\(\Rightarrow\frac{98}{33}.x+\frac{49}{33}=x+\frac{1}{33}\)

\(\Rightarrow\frac{98}{33}.x-x=\frac{1}{33}-\frac{49}{33}\)

\(\Rightarrow\frac{65}{33}.x=\frac{-16}{11}\)

\(\Rightarrow x=\frac{-16}{11}:\frac{65}{33}\)

\(\Rightarrow x=\frac{-48}{65}\)

Vậy \(x=\frac{-48}{65}\)

\(E=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)

\(E=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)

\(E=\frac{1}{1}-\frac{1}{99}\)

\(E=\frac{98}{99}\)

E= \(\frac{2}{1.3}.\frac{2}{3.5}+...+\frac{2}{97.99}\)
E = 1 - \(\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)
E = 1 - 1/99
E = 98 / 99
Chúc bạn học tốt 

đề có sai ko nhỉ ???

2Q = 1-1/3-1/2+1/4+1/3-1/5-1/4+1/6-........+1/97-1/99-1/98+1/100 = 1-1/2-1/99+1/100 = 4949/9900 >> Q = 49499/19800 

\(Q=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+...+\frac{1}{97.99}-\frac{1}{98.100}\)

\(=\frac{1}{2}\left(1-\frac{1}{3}-\frac{1}{2}+\frac{1}{4}+\frac{1}{3}+\frac{1}{5}-\frac{1}{4}+\frac{1}{6}+...+\frac{1}{97}-\frac{1}{99}-\frac{1}{98}+\frac{1}{100}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{100}\right)=\frac{1}{2}.\frac{99}{100}=\frac{99}{200}\) (không chắc cho lắm :v)