\(=3uv^2-\frac{1}{5}uv^2-367\frac{1}{4}uv^2-\frac{19}{5}uv^2+317\frac{1}{4}uv^2\)

\(=3uv^2-\frac{1}{5}uv^2-\frac{19}{5}uv^2=3uv^2-4uv^2=-uv^2\)

\(3uv^2-\left(\frac{1}{5}uv^2+367\frac{1}{4}uv^2\right)+\left(\frac{-19}{5}uv^2\right)+\left(367\frac{1}{4}uv^2\right)\)

=\(3uv^2-\frac{1}{5}uv^2-367\frac{1}{4}uv^2-\frac{19}{5}uv^2+317\frac{1}{4}uv^2\)

=\(3uv^2-\frac{1}{5}uv^2-\frac{19}{5}uv^2\)

=\(3uv^2-4uv^2\)

=\(-uv^2\)

G = \(\frac{2^2}{1.3}\).\(\frac{3^2}{2.4}\).\(\frac{4^2}{3.5}\).....\(\frac{50^2}{49.51}\)                         

=> G = \(\frac{2.2}{1.3}\).\(\frac{3.3}{2.4}\).\(\frac{4.4}{3.5}\).....\(\frac{50.50}{49.51}\)

=> G = \(\frac{2.2.3.3.4.4.....50.50}{1.2.3.3.4.4.....50.51}\)

=> G = \(\frac{2.50}{1.51}\)

=> G = \(\frac{100}{51}\)

公关稿黄继线长旧款您

\(C=\frac{5}{2}\cdot\frac{7}{5}\cdot\frac{9}{7}\cdot\frac{11}{9}\cdot...\cdot\frac{2017}{2015}\cdot\frac{2019}{2017}=\frac{2019}{2}\)

\(D=\left(1-\frac{1}{\frac{2\cdot3}{2}}\right)\cdot\left(1-\frac{1}{\frac{3\cdot4}{2}}\right)\cdot\left(1-\frac{1}{\frac{4\cdot5}{2}}\right)\cdot\left(1-\frac{1}{\frac{5\cdot6}{2}}\right)\cdot...\cdot\left(1-\frac{1}{\frac{39\cdot40}{2}}\right)\)

\(=\left(1-\frac{2}{2\cdot3}\right)\cdot\left(1-\frac{2}{3\cdot4}\right)\cdot\left(1-\frac{2}{4\cdot5}\right)\cdot\left(1-\frac{2}{5\cdot6}\right)\cdot...\cdot\left(1-\frac{2}{39\cdot40}\right)\cdot\)

Nhận xét: \(1-\frac{2}{n\left(n+1\right)}=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)nên:

\(D=\frac{4\cdot1}{2\cdot3}\cdot\frac{5\cdot2}{3\cdot4}\cdot\frac{6\cdot3}{4\cdot5}\cdot\frac{7\cdot4}{5\cdot6}\cdot\frac{8\cdot5}{6\cdot7}\cdot...\cdot\frac{41\cdot38}{39\cdot40}=\)

\(D=\frac{4\cdot5\cdot6\cdot7\cdot...\cdot41\times1\cdot2\cdot3\cdot4\cdot...\cdot38}{2\cdot3\cdot4\cdot5\cdot...\cdot39\times3\cdot4\cdot5\cdot6\cdot..\cdot40}=\frac{1}{39}\cdot\frac{41}{3}=\frac{41}{117}\)

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 

\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+....+\frac{1}{32}\left(1+2+3+...+32\right)\)

\(=1+\frac{1}{2}.\frac{2\left(2+1\right)}{2}+\frac{1}{3}.\frac{3\left(3+1\right)}{2}+....+\frac{1}{32}.\frac{32.\left(32+1\right)}{2}\)

\(=1+\frac{2+1}{2}+\frac{3+1}{2}+....+\frac{32+1}{2}\)

\(=1+\frac{3}{2}+\frac{4}{2}+....+\frac{33}{2}\)

\(\frac{2+3+4+....+33}{2}\)

\(=\frac{\frac{33\left(33+1\right)}{2}-1}{2}=280\)

tớ không biết đâu

\(1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+\frac{1}{4}.\left(1+2+3+4\right)+...+\frac{1}{20}.\left(1+...+20\right).\)

\(=1+\frac{3}{2}+\frac{6}{3}+\frac{10}{4}+...+\frac{210}{20}\)

\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{21}{2}\)

\(=\frac{2+3+4+5+...+21}{2}=\frac{230}{2}=115\)