Sửa đề : v 

\(S=\frac{14}{1.3}+\frac{14}{3.5}+\frac{14}{5.7}+\frac{14}{7.9}+...+\frac{14}{13.15}\)

\(S=7.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{13.15}\right)\)

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

\(S=7.\left(1-\frac{1}{15}\right)\)

\(S=7.\frac{14}{15}=\frac{98}{15}\)

mình tính cái đa thức ở sau nhé:

=14.(1/1.3+1/3.5+...+1/13.15)

=7.(1-1/3+1/3-1/5+...+1/13-1/15)

=7.(1-1/15)

=7.(14/15)

=98/15

còn a,b là gì thì mình ko bt

HGH̃TUYGHUGŨYHJGGBGJ

a, Ta có:

\(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{0,6-\frac{3}{9}+\frac{3}{11}}+\frac{\frac{2}{3}+\frac{2}{7}-\frac{1}{14}}{-1-\frac{3}{7}+\frac{3}{28}}=\frac{2\left(0,2-\frac{1}{9}+\frac{1}{11}\right)}{3\left(0,2-\frac{1}{9}+\frac{1}{11}\right)}+\frac{2\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{28}\right)}{-3\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{28}\right)}=\frac{2}{3}+\frac{-2}{3}=0\)

k đúng cho mình nha. Thanks!!!

a, bày cho mình cách viết bằng phân số đi , mình trình bày cách làm cho. k đúng cho mình nha.

\(=\frac{2}{3}\times\frac{0,1-\frac{1}{9}+\frac{1}{11}}{0,1-\frac{1}{9}+\frac{1}{11}}+\frac{-\frac{2}{3}\times\left(-1-\frac{3}{7}+\frac{3}{28}\right)}{-1-\frac{3}{7}+\frac{3}{28}}\) 

=\(\frac{2}{3}+\left(-\frac{3}{2}\right)\)

=\(-\frac{5}{6}\)

\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{19\cdot21}-\frac{x}{14}=\frac{2}{-7}\)

\(\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{19\cdot21}\right)-\frac{x}{14}=\frac{2}{-7}\)

\(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)-\frac{x}{14}=\frac{2}{-7}\)

\(\frac{1}{2}\left(1-\frac{1}{21}\right)-\frac{x}{14}=\frac{2}{-7}\)

\(\frac{1}{2}\cdot\frac{20}{21}-\frac{x}{14}=\frac{2}{-7}\)

\(\frac{10}{21}-\frac{x}{14}=\frac{2}{-7}\)

\(\frac{x}{14}=\frac{10}{21}-\frac{2}{-7}\)

\(\frac{x}{14}=\frac{16}{21}\)

\(\Rightarrow x\cdot=21=14\cdot16\)

\(\Rightarrow x\cdot21=224\)

\(\Rightarrow x=\frac{224}{21}\)

\(a)\) \(A=\frac{1}{199}-\frac{1}{199.198}-\frac{1}{198.197}-\frac{1}{197.196}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(A=\frac{1}{199}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{197.198}+\frac{1}{198.199}\right)\)

\(A=\frac{1}{199}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{197}-\frac{1}{198}+\frac{1}{198}-\frac{1}{199}\right)\)

\(A=\frac{1}{199}-\left(1-\frac{1}{199}\right)\)

\(A=\frac{1}{199}-1+\frac{1}{199}\)

\(A=\frac{-197}{199}\)

Chúc bạn học tốt ~ 

làm hộ mk lun câu b ik

a) \(A=\frac{1}{199}-\frac{1}{199.198}-\frac{1}{198.197}-\frac{1}{197.196}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(A=\frac{1}{199}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...\frac{1}{197.198}+\frac{1}{198.199}\right)\)

\(A=\frac{1}{199}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{197}-\frac{1}{198}+\frac{1}{198}-\frac{1}{199}\right)\)

\(A=\frac{1}{199}-(1-\frac{1}{199})\)

\(A=\frac{1}{199}-1+\frac{1}{199}\)

\(A=\frac{-197}{199}\)

Vậy \(A=\frac{-197}{199}\)

b) cũng thế 

lồn

Ta có:

\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\)

\(=\frac{1}{4}+\left(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\right)\)

Đặt \(B=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\)

\(B=\left(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}\right)+\left(\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\right)\)

Giả sử tất cả các số hạng của B đều bằng \(\frac{1}{6^2}\)

\(\Rightarrow B=6.\frac{1}{6^2}=\frac{6}{36}=\frac{1}{6}<\frac{1}{4}\)

Do đó \(B<\frac{1}{4}\)

\(\Rightarrow A=\frac{1}{4}+B<\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\)

Vậy \(A<\frac{1}{2}\)

Mình làm như thế này nek

\(\frac{\frac{1}{4}+\frac{3}{7}-\frac{4}{5}}{0,75+\frac{9}{7}-2\frac{2}{5}}+\frac{\frac{3}{14}-\frac{2}{10}+\frac{5}{18}+\frac{7}{66}}{\frac{6}{7}-\frac{4}{5}+\frac{10}{9}+\frac{14}{33}}\)

\(=\frac{\frac{1}{4}+\frac{3}{7}-\frac{4}{5}}{\frac{2}{4}+\frac{9}{7}-\frac{12}{5}}+\frac{\frac{1}{2}\cdot\left(\frac{3}{7}-\frac{2}{5}+\frac{5}{9}+\frac{7}{33}\right)}{2\cdot\left(\frac{3}{7}-\frac{2}{5}+\frac{5}{9}+\frac{7}{33}\right)}\)

\(=\frac{\frac{1}{4}+\frac{3}{7}-\frac{4}{5}}{3\cdot\left(\frac{1}{4}+\frac{3}{7}-\frac{4}{5}\right)}+\frac{\frac{1}{2}}{2}\)

\(=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)