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Ta có :
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{2016}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2015}{2016}\)
\(A=\frac{2.3.4.....2015}{2.3.4.....2015}.\frac{1}{2016}\)
\(A=\frac{1}{2016}\)
Vậy \(A=\frac{1}{2016}\)
Chúc bạn học tốt ~
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)..\left(1-\frac{1}{2016}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2015}{2016}\)
\(\Rightarrow A=\frac{1.2.3..2015}{2.3.4..2016}\)
\(\Rightarrow A=\frac{1}{2016}\)
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2016}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2015}{2016}\)
\(=\frac{1.2.3....2015}{2.3.4....2016}\)
\(=\frac{1}{2016}\)
\(\Rightarrow2A=1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{2014}\)
\(\Rightarrow2A-A=A=1-\left(\frac{1}{2}\right)^{2015}\)
Với B tương tự nhưng là lấy 3B
\(M=\frac{18.\frac{19}{2}.\frac{20}{3}...\frac{36}{19}}{20.\frac{21}{2}.\frac{22}{3}...\frac{36}{17}}=\frac{\frac{18.19.20...36}{2.3...19}}{\frac{20.21.22...36}{2.3...17}}=\frac{\frac{18.19}{18.19}}{1}=\frac{1}{1}=1\)