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\(\frac{1}{2}.\left(1+\frac{1}{1.3}\right).\left(\frac{1}{2.4}+1\right).....\left(1+\frac{1}{2015.2017}\right)\)
\(=\frac{1}{2}.\left(\frac{4}{1.3}.\frac{9}{2.4}......\frac{4064256}{2015.2017}\right)\)
\(=\frac{1}{2}.\left(\frac{2.2}{1.3}.\frac{3.3}{2.4}.....\frac{2016.2016}{2015.2017}\right)\)
\(=\frac{1}{2}.\left(\frac{2.3....2016}{1.2....2015}.\frac{2.3.....2016}{3.4....2017}\right)\)
\(=\frac{1}{2}.2016.\frac{2}{2017}\)
\(=1008.\frac{2}{2017}=\frac{2016}{2017}\)
#)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)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...0...\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
