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(3/429 - 1/1.3)(3/429 - 1/3.5) ... (3/429 - 1/121.123)
= (1/143 - 1/1.3)(1/143 - 1/3.5) ... (1/143 - 1/11.13) ... (1/143 - 1/121.123)
= (1/11.13 - 1/1.3)(1/11.13 - 1/3.5) ... (1/11.13 -1/11.13) ... (1/11.13 - 1/121.123)
= (1/11.13 - 1/1.3)(1/11.13 - 1/3.5) ... 0 ... (1/11.13 - 1/121.123)
= 0
=(1/143-1/1.3)...(1/143-1/121.123)
vì trong tích có thừa số (1/143-1/11.13)=0
nên cả tích =0
LÀM ƠN LIKE CHO MÌNH ĐI
(3/429 - 1/1.3)(3/429 - 1/3.5) ... (3/429 - 1/121.123)
= (1/143 - 1/1.3)(1/143 - 1/3.5) ... (1/143 - 1/11.13) ... (1/143 - 1/121.123)
= (1/11.13 - 1/1.3)(1/11.13 - 1/3.5) ... (1/11.13 -1/11.13) ... (1/11.13 - 1/121.123)
= (1/11.13 - 1/1.3)(1/11.13 - 1/3.5) ... 0 ... (1/11.13 - 1/121.123)
= 0
\(T=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(T=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)
\(T=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(T=2.\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(T=2.\frac{502}{1005}=\frac{1004}{1005}\)
\(\Rightarrow T=\frac{1004}{1005}\)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009+2011}\)
\(A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2009+2011}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)
\(A=\frac{1}{2}.\frac{2010}{2011}\)
\(\Rightarrow A=\frac{1005}{2011}\)