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\(C=\frac{1}{3}.\frac{1}{7}+\frac{1}{7}.\frac{1}{11}+\frac{1}{11}.\frac{1}{13}+...+\frac{1}{2011}.\frac{1}{2015}\)
\(C=\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.13}+...+\frac{1}{2011.2015}\)
\(4C=4\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.13}+...+\frac{1}{2011.2015}\right)\)
\(4C=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.13}+...+\frac{4}{2011.2015}\)
\(4C=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+...+\frac{1}{2011}-\frac{1}{2015}\)
\(4C=\frac{1}{3}-\frac{1}{2015}=\frac{2012}{6045}\)
\(C=\frac{2012}{6045}:4=\frac{503}{6045}\)
\(C=\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+....+\dfrac{1}{2011}-\dfrac{1}{2015}\right)\)
\(C=\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{2015}\right)=\dfrac{1}{4}\left(\dfrac{2012}{3.2015}\right)=\dfrac{503}{3.2015}\)
\(S=1+2+3-4-5+6+7+8-9-10+...+2011+2012+2013-2014-2015\)
\(=\left(1+2+3-4-5\right)+\left(6+7+8-9-10\right)+...+\left(2011+2012+2013-2014-2015\right)\)
\(=\left(-3\right)+2+...+2007\)
Từ 2 đến 2007 sẽ có: \(\dfrac{2007-2}{5}+1=402\left(số\right)\)
Tổng của dãy số 2;7;12;...;2007 sẽ là:
\(\dfrac{\left(2007+2\right)\cdot402}{2}=403809\)
=>S=403809-3=403806
Đây này má Ran mori
a) \(\left(5\dfrac{1}{7}-3\dfrac{3}{11}\right)-2\dfrac{1}{7}-1\dfrac{8}{11}\)
\(=5+\dfrac{1}{7}-3-\dfrac{3}{11}-2-\dfrac{1}{7}-1-\dfrac{8}{11}\)
\(=\left(5-3-2-1\right)+\left(\dfrac{1}{7}-\dfrac{3}{11}-\dfrac{1}{7}-\dfrac{8}{11}\right)\)
\(=-1+\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-\left(\dfrac{3}{11}+\dfrac{8}{11}\right)\)
\(=-1+0-1=-2\)
a)\(\left(5\dfrac{1}{7}-3\dfrac{3}{11}\right)-2\dfrac{1}{7}-1\dfrac{8}{11}\)
= \(\left(5+\dfrac{1}{7}-3+\dfrac{3}{11}\right)-2+\dfrac{1}{7}-1+\dfrac{8}{11}\)
= \(5-\dfrac{1}{7}+3-\dfrac{3}{11}-2+\dfrac{1}{7}-1+\dfrac{8}{11}\)
= \(\left(5-3-2-1\right)+\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{8}{11}-\dfrac{3}{11}\)
= \(-1+2+\dfrac{5}{11}\)
= \(1+\dfrac{5}{11}=\dfrac{1}{1}+\dfrac{5}{11}=\dfrac{11}{11}+\dfrac{5}{11}=\dfrac{16}{11}\)
Vậy :câu a) = \(\dfrac{16}{11}\)
\(\frac{5^{11}.7^{12}+5^{11}.7^{11}}{5^{12}.7^{11}+9.5^{11}.7^{11}}\)
\(=\frac{5^{11}.7^{11}.7^1+5^{11}.7^{11}}{5^{11}.5^1.7^{11}+3^2.5^{11}.7^{11}}\)
\(=\frac{5^{11}.\left(7^{11}+7^{11}\right).7^1}{5^{11}.\left(7^{11}+7^{11}\right).5^1.3^2}\)
\(=\frac{7}{5.3^2}\)
\(=\frac{7}{45}\)