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Sửa:
\(B=\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12.17}+\frac{5}{17.22}+\frac{5}{22.27}\)
Trả lời
\(B=\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12\cdot17}+\frac{5}{17\cdot22}+\frac{5}{22\cdot27}\)
\(\Rightarrow B=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{22}+\frac{1}{22}-\frac{1}{27}\)
\(\Rightarrow B=\frac{1}{2}-\frac{1}{27}\)
\(\Rightarrow B=\frac{25}{54}\)
Vậy B=\(\frac{25}{54}\)
\(\frac{5}{2\cdot7}+\frac{5}{7\cdot12}+\frac{5}{12\cdot17}+\frac{5}{17\cdot22}+\frac{5}{22\cdot27}\)
\(=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{22}+\frac{1}{22}-\frac{1}{27}\)
\(=\frac{1}{2}-\frac{1}{27}\)
\(=\frac{25}{54}\)
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!
\(\frac{5}{2}-\frac{5}{7}+\frac{5}{7}-\frac{5}{12}+\frac{5}{12}-\frac{5}{17}+\frac{5}{17}-\frac{5}{22}+\frac{5}{22}-\frac{5}{29}=\frac{5}{2}-0-0-0-0-\frac{5}{29}=\frac{5}{2}-\frac{5}{29}=\frac{145}{58}-\frac{10}{58}=\frac{135}{58}\)
Đặt
\(A=\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12.17}+\frac{5}{17.22}+\frac{5}{22.27}+\frac{5}{27.32}+\frac{5}{32.37}\)
\(A=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{22}+\frac{1}{22}-\frac{1}{27}+\frac{1}{27}-\frac{1}{32}+\frac{1}{32}-\frac{1}{37}\)
\(A=\frac{1}{2}-\frac{1}{37}=\frac{35}{74}\)
\(=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{32}-\frac{1}{37}\)
\(=\frac{1}{2}-\frac{1}{37}\)
\(=\frac{37}{74}-\frac{2}{74}=\frac{35}{74}\)
\(\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12.17}+\frac{5}{22.27}\)
\(=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{5}{22.27}\)
\(=\frac{1}{2}-\frac{1}{17}+\frac{5}{22.27}\)
\(=\frac{2270}{5049}\)
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 = \(2015+\frac{10}{2,7}+\frac{10}{7,12}+\frac{10}{12,17}+\frac{10}{17,22}=2021.510611\)
Chỉ cần để các thừa số ra ngoài rồi nhân các số mà bằng khoảng cách của mẫu lên tử là giải được
\(S2=\frac{5}{2\cdot7}+\frac{5}{7\cdot12}+\frac{5}{12\cdot17}+\frac{5}{17\cdot22}+\frac{5}{22\cdot27}\)
\(S2=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{22}+\frac{1}{22}-\frac{1}{27}\)
\(S2=\frac{1}{2}-\frac{1}{27}\)
\(S2=\frac{52}{54}=\frac{26}{27}\)
CHÚC BN HC TỐT !!!!!!!!! TỨ DIỆP THẢO
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