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

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

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

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:

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

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

\(\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)...\left(1+\frac{1}{2004\cdot2006}\right)\)

\(=\frac{4}{1\cdot3}+\frac{9}{2\cdot4}+\frac{16}{3\cdot5}+...+\frac{420025}{2004\cdot2006}\)

\(=\frac{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2005\cdot2005\right)}{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2004\cdot2006\right)}\)

\(=\frac{\left(2\cdot3\cdot4\cdot...\cdot2005\right)\left(2\cdot3\cdot4\cdot...\cdot2005\right)}{\left(1\cdot2\cdot3\cdot...\cdot2004\right)\left(3\cdot4\cdot5\cdot...\cdot2006\right)}\)

\(=\frac{2005\cdot2}{1\cdot2006}\)

\(=\frac{4010}{2006}\)

\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)...\left(1+\frac{1}{2004.2006}\right)\)

\(=\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}....\frac{2004.2006+1}{2004.2006}\)

\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}....\frac{2005^2}{2004.2006}\)

\(=\frac{2.3....2005}{1.2....2004}.\frac{2.3...2005}{3.4....2006}\)

\(=2005.\frac{2}{2006}=\frac{2005}{1003}\)

\(=\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right)...\left(1+\frac{1}{2004.2006}\right)\)

\(=\left(\frac{2.2}{1.3}\right).\left(\frac{3.3}{2.4}\right)....\left(\frac{2005.2005}{2004.2006}\right)\)

\(=\frac{\left(2.3.4....2005\right).\left(2.3.4....2005\right)}{\left(1.2.3...2004\right).\left(3.4.5....2006\right)}\)

\(=\frac{2005}{2006}\)

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

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

\(=\frac{\left(2.3.4.....50\right).\left(2.3.4.....50\right)}{\left(1.2.3.....49\right).\left(3.4.5.....51\right)}\)

\(=\frac{50.2}{51}=\frac{100}{51}\)

\(H=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right).\left(1-\frac{3}{7}\right).....\left(1-\frac{10}{7}\right)\)

\(=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right).\left(1-\frac{3}{7}\right).....\left(1-\frac{7}{7}\right).....\left(1-\frac{10}{7}\right)\)

\(=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right).\left(1-\frac{3}{7}\right).....0.....\left(1-\frac{10}{7}\right)\)

\(=0\)

đụ cha mi

mi trù ta thi rớt HK II mà ta giúp mày hả

mấy bài này cũng dễ ẹt nữa

đừng có mơ ta sẽ giúp mày

ha ha ha ha ha ha ha ha ha ha ha ha ha ha ha ha
 

\(B=\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)...\left(1+\frac{1}{99\cdot101}\right)\)

\(B=\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot5}\cdot\cdot\cdot\frac{100^2}{99\cdot101}\)

\(B=\frac{2^2\cdot3^2\cdot4^2\cdot\cdot\cdot100^2}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot\cdot\cdot99\cdot101}\)

\(B=\frac{\left(2\cdot3\cdot4\cdot\cdot\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot\cdot\cdot100\right)}{\left(1\cdot2\cdot3\cdot\cdot\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot\cdot\cdot101\right)}\)

\(B=\frac{100\cdot2}{1\cdot101}\)

\(B=\frac{200}{101}\)