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a=1/3x5+1/5x7+...+1/2003x2005
a=1x2/3x5x2+1x2/5x7x2+...+1x2/2003x2005x2
a=1/2(2/3x5+2/5x7+...+2/2003x2005)
a=1/2x(1/3-1/5+1/5-1/7+...+1/2003-1/2005)
a=1/2x(1/3-1/2005)
a=1/2x2002/6015
a=1001/6015
a. \(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2014}}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{2013}}\)
\(\Rightarrow3A-A=1-\frac{1}{3^{2014}}\)
\(\Rightarrow2A=1-\frac{1}{3^{2014}}\)
\(\Rightarrow A=\left(1-\frac{1}{3^{2014}}\right):2=\frac{1}{2}-\frac{1}{3^{2014}.2}=\frac{3^{2014}-1}{3^{2014}.2}\)
b.\(B=\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2014}}\)
\(\Rightarrow2B=1+\frac{1}{2^2}+....+\frac{1}{2^{2013}}\)
\(\Rightarrow2B-B=1-\frac{1}{2^{2014}}\)
\(\Rightarrow B=1-\frac{1}{2^{2014}}\)
a) \(A=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+......+\frac{1}{2017.2022}\)
\(5A=5.\left(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+.....+\frac{1}{2017.2022}\right)\)
\(5A=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+......+\frac{5}{2017.2022}\)
\(5A=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+........+\frac{1}{2017}-\frac{1}{2022}\)
\(5A=1-\frac{1}{2022}\)
\(5A=\frac{2022}{2022}-\frac{1}{2022}\)
\(5A=\frac{2021}{2022}\)
\(A=\frac{2021}{2022}\div5\)
\(A=\frac{20201}{10110}\)
TL:
\(\frac{5}{6}=\frac{1}{2}+\frac{1}{3}\)
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HT
đề này hỏi j