D = \(\frac{1}{54}-\frac{3}{1.3}-\frac{3}{3.5}-\frac{3}{5.7}-...-\frac{1}{79.81}\)

\(=\frac{1}{54}-\left(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{79.81}\right)\)

\(=\frac{1}{54}-\frac{3}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{79.81}\right)\)

\(=\frac{1}{54}-\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{79}-\frac{1}{81}\right)\)

\(=\frac{1}{54}-\frac{3}{2}.\left(1-\frac{1}{81}\right)\)

\(=\frac{1}{54}-\frac{3}{2}.\frac{80}{81}\)

\(=\frac{1}{54}-\frac{40}{27}\)

\(=\frac{1}{54}-\frac{80}{54}\)

\(=\frac{79}{54}\)

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=3(1/1.3+1/3.5+1/5.7+1/7.9)

=3/2(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9) vi khoang cach tu 1-3;3-5;5-7;7-9 la 2 nen ta nhan tat ca voi 1/2 ma 3.1/2=3/2

=3/2.(1-1/9) rut gon -1/3+1/3;-1/5+1/5;-1/7+1/7=0

=3/2.8/9=4/3

ta có :3/(1.3)+3/(3.5)+3/(5.7)+3/(7.9)

ta đặt 3 làm chung rồi tự làm đc

dat 3 lam chung

A có tổng cộng 49 số hạng, nhóm 2 số hạng liên tiếp với nhau được: 

\(A=\left(\frac{1}{1.3}-\frac{2}{3.5}\right)+\left(\frac{3}{5.7}-\frac{4}{7.9}\right)+...+\left(\frac{47}{93.95}-\frac{48}{95.97}\right)+\frac{49}{97.99}\)

\(A=\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{93.97}+\frac{49}{97.99}\)=> \(4A=\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{93.97}+\frac{196}{97.99}=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{93}-\frac{1}{97}+\frac{196}{97.99}\)

=> \(4A=1-\frac{1}{97}+\frac{196}{97.99}=\frac{96}{97}+\frac{196}{97.99}=\frac{9700}{97.99}=\frac{100}{99}>1\)

\(4A>1=>A>\frac{1}{4}\)

Bn trừ 2 PS kiểu gì hay zậy? 

Giúp mình nhá

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

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

\(=\frac{1}{3}-\frac{1}{11}=\frac{11}{33}-\frac{3}{33}=\frac{8}{33}\)

a)(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{11.12}\)). x=\(\frac{1}{3}\)

(1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{11}_{ }+\frac{1}{12}\)).x=\(\frac{1}{3}\)

(1+\(\frac{1}{12}\)).x=\(\frac{1}{3}\)

x=\(\frac{1}{3}:\frac{13}{12}\)

x=\(\frac{4}{13}\)

b)( \(2-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+...+\frac{2}{9}-\frac{2}{11}_{ }\)):x =\(\frac{2}{3}\)

Giống câu a

=> \(\Rightarrow\left(\frac{11}{5}-\frac{11}{7}+\frac{11}{7}-\frac{11}{9}+...+\frac{11}{59}-\frac{11}{61}\right):2=\left(\frac{11}{5}-\frac{11}{61}\right):2=\frac{616}{305}:2=\frac{308}{305}\)

Đặt \(A=\frac{11}{5.7}+\frac{11}{7.9}+...+\frac{11}{59.61}\)

\(\Rightarrow2A:11=\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\)

\(\Rightarrow2A:11=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\)

\(\Rightarrow2A:11=\frac{1}{5}-\frac{1}{61}\)

\(\Rightarrow2A:11=\frac{56}{305}\)

\(\Rightarrow2A=\frac{56}{305}.11=\frac{616}{305}\)

\(\Rightarrow A=\frac{616}{305}:2=\frac{308}{305}\)

Vậy kết quả của phép tính trên là \(\frac{308}{305}\)

11/2x(1/5-1/7+1/7+1/9+.....+1/59-1/61)

=11/2x(1/5-1/61)

=11/2x56/305

=308/305