\(\frac{3}{2.7}+\frac{3}{7.12}+\frac{3}{12.17}+...+\frac{3}{2012.2017}\)
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Ta có : 51 + 52 +...+ 52016
= (5+52+53)+(54+55+56)+...+(52014+52015+52016)
= 5(1+5+52)+53(1+5+52)+...+52013(1+5+52)
= 5. 31 + 53.31 +...+52013.31
= 31(5+52+...+52013) chia hết cho 31
Vây 51 + 52 + ...+ 52016 chia hết cho 31
Chỉ cần để các thừa số ra ngoài rồi nhân các số mà bằng khoảng cách của mẫu lên tử là giải được
Cho S=\(\frac{5}{2.7}+\frac{5}{7.12}+...+\frac{5}{22.27}\)
\(=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{22}-\frac{1}{27}\)
\(=\frac{1}{2}-\frac{1}{27}=\frac{25}{54}\)
kb nha mn!
\(\frac{5}{2}-\frac{5}{7}+\frac{5}{7}-\frac{5}{12}+\frac{5}{12}-\frac{5}{17}+\frac{5}{17}-\frac{5}{22}+\frac{5}{22}-\frac{5}{29}=\frac{5}{2}-0-0-0-0-\frac{5}{29}=\frac{5}{2}-\frac{5}{29}=\frac{145}{58}-\frac{10}{58}=\frac{135}{58}\)
TA CÓ: \(\frac{1}{n}-\frac{1}{n+5}=\frac{n+5-n}{n\left(n+5\right)}=\frac{5}{n\left(n+5\right)}\)
Thay vào biểu thức trên , ta được:
\(\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12.17}+...+\frac{5}{907.1002}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{907}-\frac{1}{1002}\)
\(=\frac{1}{2}-\frac{1}{1002}=\frac{250}{501}\)
\(N=2015+\frac{10}{2.7}+\frac{10}{7.12}+\frac{10}{12.17}+\frac{10}{17.22}\)
\(=2\left(1007,5+\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12.17}+\frac{5}{17.22}\right)\)
\(=2\left(1007,5+\frac{7-2}{2.7}+\frac{12-7}{7.12}+\frac{17-12}{12.17}+\frac{22-17}{17.22}\right)\)
\(=2\left(1007,5+\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{22}\right)\)
\(=2\left(1007,5+\frac{1}{2}-\frac{1}{22}\right)\)
\(=2015+1-\frac{1}{11}\)
\(=\frac{22175}{11}\)
N = \(2015+\frac{10}{2,7}+\frac{10}{7,12}+\frac{10}{12,17}+\frac{10}{17,22}=2021.510611\)
Đặt
\(A=\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12.17}+\frac{5}{17.22}+\frac{5}{22.27}+\frac{5}{27.32}+\frac{5}{32.37}\)
\(A=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{22}+\frac{1}{22}-\frac{1}{27}+\frac{1}{27}-\frac{1}{32}+\frac{1}{32}-\frac{1}{37}\)
\(A=\frac{1}{2}-\frac{1}{37}=\frac{35}{74}\)
\(=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{32}-\frac{1}{37}\)
\(=\frac{1}{2}-\frac{1}{37}\)
\(=\frac{37}{74}-\frac{2}{74}=\frac{35}{74}\)
\(\frac{1}{2.7}+\frac{1}{7.12}+\frac{1}{12.17}+....+\frac{1}{2012.2017}\)
\(=\frac{1}{5}\left(\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12.17}+....+\frac{5}{2012.2017}\right)\)
\(=\frac{1}{5}\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+....+\frac{1}{2012}-\frac{1}{2017}\right)\)
\(=\frac{1}{5}\left(\frac{1}{2}-\frac{1}{2017}\right)\)
\(=\frac{1}{5}.\frac{2015}{4034}=\frac{403}{4034}\)
ĐẶT A=DÃY SỐ TRÊN=>5A=5/2.7+........+5/2012.2017
=>A=1/2-1/7........-1/2012-1/2017 RÚT GỌN TA ĐƯỢC A=1/2-1/2017
\(B=\frac{3}{2.7}+\frac{3}{7.12}+\frac{3}{12.17}+...+\frac{3}{87.92}\)
\(\Leftrightarrow\frac{5}{3}B=\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12.17}+...+\frac{5}{87.92}\)
\(\Leftrightarrow\frac{5}{3}B=\frac{7-2}{2.7}+\frac{12-7}{7.12}+\frac{17-12}{12.17}+...+\frac{92-87}{87.92}\)
\(\Leftrightarrow\frac{5}{3}B=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{87}-\frac{1}{92}\)
\(\Leftrightarrow\frac{5}{3}B=\frac{1}{2}-\frac{1}{92}\)
\(\Leftrightarrow\frac{5}{3}B=\frac{46}{92}-\frac{1}{92}\)
\(\Leftrightarrow\frac{5}{3}B=\frac{45}{92}\)
\(\Leftrightarrow B=\frac{45}{92}\times\frac{3}{5}\)
\(\Leftrightarrow B=\frac{27}{92}\)
CTV làm màu vc :)
\(B=\frac{3}{2.7}+\frac{3}{7.12}+\frac{3}{12.17}+...+\frac{3}{87.92}\)
\(B=\frac{3}{5}.\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{87}-\frac{1}{92}\right)\)
\(B=\frac{3}{5}.\left(\frac{1}{2}-\frac{1}{92}\right)\)
\(B=\frac{3}{5}.\frac{45}{92}\)
\(B=\frac{27}{92}\)
\(\Rightarrow C=\frac{10}{5}\left(\frac{1}{7.12}+\frac{1}{12.17}+\frac{1}{17.22}+...+\frac{1}{502.507}\right)\)
\(\Rightarrow C=2.\left(\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+....+\frac{1}{507}-\frac{1}{507}\right)\)
\(\Rightarrow C=2.\left(\frac{1}{7}-\frac{1}{507}\right)=2.\frac{1}{7}-2.\frac{1}{507}=\frac{2}{7}-\frac{2}{507}\)
Đặt A=đã cho
=>\(\frac{5}{3}A=\frac{5}{2\cdot7}+\frac{5}{7\cdot12}+...+\frac{5}{2012\cdot2017}\)
=>\(\frac{5}{3}A=\frac{1}{2}-\frac{1}{2017}\)
Đến đây dễ rồi tự lm tiếp nhé
\(=\frac{3}{5}.\left(\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12.17}+...+\frac{5}{2012.2017}\right)\)
\(=\frac{3}{5}.\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{2012}-\frac{1}{2017}\right)\)
\(=\frac{3}{5}.\left(\frac{1}{2}-\frac{1}{2017}\right)\)
\(=\frac{3}{5}.\left(\frac{2015}{4034}\right)\)
\(=\frac{1209}{4034}\)
Bài này bạn phải học lý thuyết mới làm được nhé!! Chúc bạn zui~^^