A = \(\left(\frac{1}{8}+\frac{1}{8.15}+\frac{1}{15.22}+...+\frac{1}{43.50}\right)\cdot\frac{4-3-5-7-...-49}{217}\)

A = \(\frac{1}{7}.\left(\frac{7}{1.8}+\frac{7}{8.15}+\frac{7}{15.22}+...+\frac{7}{43.50}\right)\cdot\frac{4-\left(3+5+7+...+49\right)}{217}\)

A = \(\frac{1}{7}.\left(1-\frac{1}{8}+\frac{1}{8}-\frac{1}{15}+\frac{1}{15}-\frac{1}{22}+...+\frac{1}{43}-\frac{1}{50}\right)\cdot\frac{4-\left(49+3\right)\left[\left(49-3\right):2+1\right]:2}{217}\)

A = \(\frac{1}{7}\cdot\left(1-\frac{1}{50}\right)\cdot\frac{4-52.24:2}{217}\)

A = \(\frac{1}{7}\cdot\frac{49}{50}\cdot\frac{4-624}{217}\)

A = \(\frac{7}{50}\cdot\frac{-620}{217}=-\frac{2}{5}\)

a)\(VT=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)

\(=\frac{1}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right]\)

\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right]\)

\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{3n+2}\right]=\frac{1}{3}\left[\frac{3n+2}{2\left(3n+2\right)}-\frac{2}{2\left(3n+2\right)}\right]\)

\(=\frac{1}{3}\cdot\frac{3n}{6n+4}=\frac{n}{6n+4}=VP\)

b) Ta có: \(\frac{5}{3.7}+\frac{5}{7.11}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)

\(=\frac{5}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\right)\)

\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{4n-1}-\frac{1}{4n+3}\right)\)

\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{4n+3}\right)\)

\(=\frac{5}{4}\left(\frac{4n+3}{12n+9}-\frac{3}{12n+9}\right)\)

\(=\frac{5}{4}.\frac{4n}{12n+9}\)

\(=\frac{5n}{12n+9}\)

( sai đề )

a) \(\left(\frac{11}{4}.\frac{-5}{9}-\frac{4}{9}.\frac{11}{4}\right).\frac{8}{33}\)

=\(\frac{11}{4}\left(-\frac{5}{9}-\frac{4}{9}\right).\frac{8}{33}\)

=\(\frac{11}{4}\cdot-1\cdot\frac{8}{33}\)

=\(-\frac{11}{4}\cdot\frac{8}{33}\)

=\(-\frac{2}{3}\)

b)\(-\frac{1}{4}\cdot\frac{152}{11}+\frac{68}{4}\cdot-\frac{1}{11}\)

=\(\frac{-1.152}{4.11}+\frac{68}{4}\cdot\frac{-1}{11}\)

=\(\frac{-1.152}{11.4}+\frac{68}{4}\cdot\frac{-1}{11}\)

=\(\frac{-1}{11}\cdot\frac{152}{4}+\frac{68}{4}\cdot\frac{-1}{11}\)

=\(\frac{-1}{11}\cdot\left(\frac{152}{4}+\frac{68}{4}\right)\)

=\(\frac{-1}{11}\cdot55=-5\)

c)\(\frac{-2}{3}\cdot\frac{4}{5}+\frac{2}{3}\cdot\frac{3}{5}\)

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

=\(-1\cdot\frac{2}{3}\cdot\frac{7}{5}\)

=\(-\frac{2}{3}\cdot\frac{7}{5}\)

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

d) chưa nghĩ ra nhé

e) bạn chép sai đề bài rồi

mk mới kiểm tra 45 phút nên biết

đề bài nè

\(\frac{3}{2^2}\cdot\frac{8}{3^2}\cdot\frac{15}{4^2}\cdot...\cdot\frac{899}{30^2}\)

=\(\frac{1.3}{2^2}\cdot\frac{2.4}{3^2}\cdot\frac{3.5}{4^2}\cdot...\cdot\frac{29.31}{30^2}\)

=\(\frac{1.3.2.4.3.5...29.31}{2.2.3^2.4^2...30.30}\)

=\(\frac{1.2.3^2.4^2.5^2....29^2.30.31}{2.2.3^2.4^2.5^2....29^2.30.30}\)

=\(\frac{1.31}{2.30}\)

=\(\frac{31}{60}\)

a)trong ngoac bn dat thau so chung la 11/4 rui tinh binh thuong                                                                         b)bn tu lam nhe                                                                                                                                             c)dat thua so chung                                                                                                                                       d)tinh trong ngoac ra rui nhan vs                                                                                                                       e) mk bo tay 

thanks mấy pn nha

yutyugubhujyikiu

kazuto kirigaya thật là bt làm ko đó ko bt thì nói đi còn bt thì làm đi

trời ơi bài dễ thế này tự làm đi còn hỏi

Đã biết thì đã không hỏi. Đồ kazuto kirigaya xấu tính!