Cho \(Q=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)Vậy 10Q=?
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\(Q=\frac{1}{30}+\frac{1}{42}+\frac{1}{72}+\frac{1}{90}\)
\(=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{5}-\frac{1}{10}=\frac{1}{10}\)
\(\Rightarrow10Q=\frac{1}{10}.10=1\)
\(Q\) = \(\frac{1}{3}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
\(Q\) = \(\frac{5}{14}+\frac{2}{63}+\frac{1}{90}\)
\(Q\) = \(\frac{7}{18}\)+ \(\frac{1}{90}\)
\(Q=\frac{2}{5}\)
\(10Q=\frac{2}{5}x10\)
\(10Q=4\)
Sai đề rồi
Khong phải là ..... ... . . .. . .
mà là +++++++++++
\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+.....+\frac{1}{132}\)
\(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+.....+\frac{1}{11.12}\)
\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+....+\frac{1}{11}-\frac{1}{12}\)
\(A=\frac{1}{5}-\frac{1}{12}\)
\(A=\frac{7}{60}\)
Ta có:
A = \(\frac{1}{5.6}\)+ \(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)+\(\frac{1}{8.9}\)+\(\frac{1}{9.10}\)+\(\frac{1}{10.11}\)+\(\frac{1}{11.12}\)
Bạn xem lời giải của mình nhé:
Giải:
\(A=\frac{1}{30}+\frac{1}{42}+...+\frac{1}{132}\\ =\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{11.12}\\ =\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\\ =\frac{1}{5}-\frac{1}{12}=\frac{12-5}{60}=\frac{7}{60}\)
Chúc bạn học tốt!
\(\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
\(=-\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(=-\left(\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\right)\)
\(=-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(=-\left(\frac{1}{4}-\frac{1}{10}\right)\)
\(=-\frac{3}{20}\)
Bài làm:
Ta có: \(\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
\(=-\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(=-\left(\frac{1}{4}-\frac{1}{10}\right)\)
\(=-\frac{3}{20}\)
\(A=\frac{1}{20}+\frac{1}{30}+...+\frac{1}{132}\)
\(A=\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{11\times12}\)
\(A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{11}-\frac{1}{12}\)
\(A=\frac{1}{4}-\frac{1}{12}\)
\(A=\frac{3}{12}-\frac{1}{12}=\frac{2}{12}=\frac{1}{6}\)
\(Q=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(Q=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{9}-\frac{1}{10}\)
\(Q=\frac{1}{5}-\frac{1}{10}=\frac{1}{10}\)
Vậy 10Q=1