Các câu hỏi tương tự
\(A=-\frac{1}{2}\left(17,5-7,5\right)-\frac{2015}{2016}\left(2018-2\right)\)
=> \(A=-\frac{1}{2}\left(10\right)-\frac{2015}{2016}\left(2016\right)=-5-2015=-2020\)
\(\left(\frac{1}{2}+\frac{2015}{2016}+\frac{2016}{2017}+1\right)\left(\frac{2105}{2016}+\frac{2016}{2017}+\frac{7}{22}\right)-\left(\frac{1}{2}+\frac{2015}{2016}+\frac{2016}{2017}\right)\left(\frac{2015}{2016}+\frac{2016}{2017}+\frac{7}{22}+1\right)\)
A = \(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}\)
B = 3.{ 5 . [ ( 5^2 + 2^3 ) : 11 ] -16} + 2015
C = \(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right).....\left(1+\frac{1}{2014.2016}\right)\)
1 . Tinh : a , left(1-frac{1}{6}right)left(1-frac{1}{10}right)left(1-frac{1}{14}right).....left(1-frac{1}{5050}right)b,frac{^{2^{19}}.left(3^3right)^3+3.5.left(2^2right)^9.left(3^2right)^4}{left(2.3right)^9.2^{10}+left(3.2^2right)^{10}}c,frac{18.frac{19}{2}.frac{20}{3}.frac{21}{4}.....frac{36}{19}}{20.frac{21}{2}.frac{22}{3}.....frac{36}{17}}giup mjk nha mjk tjk cho
Đọc tiếp
1 . Tinh : a , \(\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{14}\right).....\left(1-\frac{1}{5050}\right)\)b,\(\frac{^{2^{19}}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(3.2^2\right)^{10}}\)c,\(\frac{18.\frac{19}{2}.\frac{20}{3}.\frac{21}{4}.....\frac{36}{19}}{20.\frac{21}{2}.\frac{22}{3}.....\frac{36}{17}}\)giup mjk nha mjk tjk cho
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{2}{3}\right)...\left(1-\frac{2015}{2016}\right)\)
1. \(B=\frac{12}{\left(2.4\right)^2}+\frac{20}{\left(4.6\right)^2}+...+\frac{388}{\left(96.98\right)^2}+\frac{396}{\left(98.100\right)^2}\)
So sánh \(B\) với \(\frac{1}{4}\)
2. SO sánh \(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\) và \(B=\frac{2015+2016+2017}{2016+2017+2018}\)
Tính A = \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2014}\right)\left(1-\frac{1}{2015}\right)\left(1-\frac{1}{2016}\right)\)
Tính S :
\(S=\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)\) \(+...+\frac{1}{2015}\left(1+2+...+2014+2015\right)\) \(+\frac{1}{2016}\left(1+2+...+2015+2016\right)\)
Tính S :
\(S=\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)\) \(+...+\frac{1}{2015}\left(1+2+...+2014+2015\right)\) \(+\frac{1}{2016}\left(1+2+...+2015+2016\right)\)