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= (1/2).(2/3).(4/5).(5/6)......(2016/2017).(2017/2018)
=1.2.3.4.5......2016.2017/2.3.4.5.....2017.2018
=1/2018
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\cdot\cdot\cdot\cdot\cdot\left(1-\frac{1}{2017}\right)\left(1-\frac{1}{2018}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\cdot\cdot\frac{2016}{2017}\cdot\frac{2017}{2018}\)
\(=\frac{1\cdot2\cdot3\cdot\cdot\cdot\cdot\cdot2016\cdot2017}{2\cdot3\cdot4\cdot\cdot\cdot\cdot2017\cdot2018}\)
\(=\frac{1}{2018}\)
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 = \(\frac{100}{51}\)
#)Giải :
a)\(2009^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-15^3\right)}=2009^{\left(1000-1^3\right)...\left(1000-10^3\right)...\left(1000-15^3\right)}=2009^0=1\)
b)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...0...\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
\(\frac{\left(\frac{2}{3}\right)^3\cdot\left(-\frac{3}{4}^2\right)\cdot\left(-1\right)^{2003}}{\left(\frac{2}{5}\right)^2\cdot\left(-\frac{5}{12}\right)^3}\)
\(=\frac{\frac{8}{27}\cdot\frac{9}{16}\cdot\left(-1\right)}{\frac{4}{25}\cdot\left(-\frac{125}{1728}\right)}\)
\(=\frac{-\frac{1}{6}}{-\frac{5}{432}}=-\frac{1}{6}:\left(-\frac{5}{432}\right)=\frac{72}{5}\)
\(\left[6.\left(\frac{-1}{3}\right)^2-3.\left(\frac{-1}{3}\right)+1\right]:\left(\frac{-1}{3}-1\right)\)
\(=\left[6.\frac{1}{9}-\left(-1\right)+1\right]:\frac{-4}{3}\)
\(=\left[\frac{2}{3}-\left(-1\right)+1\right]:\frac{-4}{3}\)
\(=\frac{8}{3}:\frac{-4}{3}=\frac{-24}{12}=-2\)
~ Hok tốt ~
\(C=\frac{5}{2}\cdot\frac{7}{5}\cdot\frac{9}{7}\cdot\frac{11}{9}\cdot...\cdot\frac{2017}{2015}\cdot\frac{2019}{2017}=\frac{2019}{2}\)
\(D=\left(1-\frac{1}{\frac{2\cdot3}{2}}\right)\cdot\left(1-\frac{1}{\frac{3\cdot4}{2}}\right)\cdot\left(1-\frac{1}{\frac{4\cdot5}{2}}\right)\cdot\left(1-\frac{1}{\frac{5\cdot6}{2}}\right)\cdot...\cdot\left(1-\frac{1}{\frac{39\cdot40}{2}}\right)\)
\(=\left(1-\frac{2}{2\cdot3}\right)\cdot\left(1-\frac{2}{3\cdot4}\right)\cdot\left(1-\frac{2}{4\cdot5}\right)\cdot\left(1-\frac{2}{5\cdot6}\right)\cdot...\cdot\left(1-\frac{2}{39\cdot40}\right)\cdot\)
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:
\(D=\frac{4\cdot1}{2\cdot3}\cdot\frac{5\cdot2}{3\cdot4}\cdot\frac{6\cdot3}{4\cdot5}\cdot\frac{7\cdot4}{5\cdot6}\cdot\frac{8\cdot5}{6\cdot7}\cdot...\cdot\frac{41\cdot38}{39\cdot40}=\)
\(D=\frac{4\cdot5\cdot6\cdot7\cdot...\cdot41\times1\cdot2\cdot3\cdot4\cdot...\cdot38}{2\cdot3\cdot4\cdot5\cdot...\cdot39\times3\cdot4\cdot5\cdot6\cdot..\cdot40}=\frac{1}{39}\cdot\frac{41}{3}=\frac{41}{117}\)