Tính giá trị biểu thức: 1/30 + 1/42 + 1/56+ 1/72+1/90+1/110+1/132
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\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{132}\)
\(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{11.12}\)
\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{11}-\frac{1}{12}\)
\(A=\frac{1}{5}-\frac{1}{12}\)
\(A=\frac{12}{60}-\frac{5}{60}=\frac{7}{60}\)
\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(\Rightarrow A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)
\(\Rightarrow A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(\Rightarrow A=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)
A=1/5x6+1/6x7+1/7x8+1/8x9+1/9x10+1/10x11+1/11x12
A=1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10+1/11-1/12
A=1/5-1/12
A=7/60
\(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{110}+\frac{1}{132}\)
\(=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{10.11}+\frac{1}{11.12}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{5}-\frac{1}{12}=\frac{12}{60}-\frac{5}{60}=\frac{7}{60}\)
A=1/30 + 1/42 + ...+ 1/132
=1/5.6 + 1/6.7 +1/7.8 +....+1/11.12
=6-5/5.6 + 7-6/6.7+ ...+ 12-11/11.12
=1/5 -1/6 +1/6 - 1/7+....+ 1/11 - 1/12
=1/5-1/12
=7/60
\(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}=\frac{6-5}{5.6}+\frac{7-6}{6.7}+...+\frac{12-11}{11.12}\)
\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)
A = 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9 + 1/9.10 + 1/10.11 + 1/11.12
= 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + ... + 1/11 - 1/12
= 1/5 - 1/12
= 12/60 - 5/60
= 7/60
Vậy A = 7/60.
Xét A , ta thấy:
\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)
Ta lại thấy: \(\frac{1}{5.6}=\frac{1}{5}-\frac{1}{6}\)
\(\frac{1}{6.7}=\frac{1}{6}-\frac{1}{7}\)
....................
\(\frac{1}{11.12}=\frac{1}{11}-\frac{1}{12}\)
\(A=\left(\frac{1}{5}-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{7}\right)+....+\left(\frac{1}{11}-\frac{1}{12}\right)\)
\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-....-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(A=\frac{1}{5}+\left(-\frac{1}{6}+\frac{1}{6}\right)+\left(-\frac{1}{7}+\frac{1}{7}\right)+....+\left(-\frac{1}{11}+\frac{1}{11}\right)-\frac{1}{12}\)
\(A=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)
\(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{11.12}=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)
\(A=\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(=\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)
\(=\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{7}-\frac{1}{12}=\frac{12}{84}-\frac{7}{84}=\frac{5}{84}\)
Vậy \(A=\frac{5}{84}\).
Bài làm:
Ta có: \(A=\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(A=\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)
\(A=\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(A=\frac{1}{7}-\frac{1}{11}\)
\(A=\frac{4}{77}\)
\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{7}{60}\)