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=1/6*7+1/7*8+1/8*9...+1/14*15
=1/6-1/7+1/7-1/8+...+1/14-1/15
=1/6-1/15
=1/10
1/42 + 1/56 + 1/72 + 1/90 + 1/110 + 1/132 + 1/156 + 1/182 + 1/210
= 1/6.7 + 1/7.8 + 1/8.9 + 1/9.10 + 1/10.11 + 1/11.12 + 1/12.13 + 1/13.14 + 1/14.15
= 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10 + 1/10 - 1/11 + 1/11 - 1/12 + 1/12 - 1/13 + 1/13 - 1/14 + 1/14 - 1/15
= 1/6 - 1/15
= 1/10
\(\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+....+\frac{1}{240}=\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+...+\frac{1}{15.16}\)
\(=\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+...+\frac{1}{15}-\frac{1}{16}=\frac{1}{7}-\frac{1}{16}=\frac{9}{112}\)
\(=\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}+\frac{1}{14.15}+\frac{1}{15.16}\)
\(=\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{16}\)
\(=\frac{1}{7}-\frac{1}{16}\)
\(=\frac{9}{112}\)
\(A=\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}+\frac{1}{156}+\frac{1}{182}+\frac{1}{210}\)
\(A=\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}+\frac{1}{14.15}\)
\(A=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}\)
\(A=\frac{1}{6}-\frac{1}{15}\)
\(A=\frac{1}{10}\)
A=\(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}+\frac{1}{156}+\frac{1}{182}+\frac{1}{210}\)
=\(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}+\frac{1}{14.15}\)
=\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+....+\frac{1}{14}-\frac{1}{15}\)
=\(\frac{1}{6}-\frac{1}{15}=\frac{1}{10}\)
\(A=\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+...+\frac{1}{210}=\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+...+\frac{1}{14.15}\)
\(=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+...+\frac{1}{14}-\frac{1}{15}\)
\(=\frac{1}{6}-\frac{1}{15}=\frac{1}{10}\)
\(A=\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}+\frac{1}{156}+\frac{1}{182}+\frac{1}{210}\)
\(A=\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}+\frac{1}{14.15}\)
\(A=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}\)
\(A=\frac{1}{6}-\frac{1}{15}\)
\(A=\frac{1}{10}\)
không lẽ nào bằng nhau được đâu vì
\(\left(\frac{1}{30}+\frac{1}{42}\right)+\left(\frac{1}{56}+\frac{1}{72}\right)+\left(\frac{1}{90}+\frac{1}{110}\right)+\left(\frac{1}{132}+\frac{1}{156}\right)+\left(\frac{1}{182}+\frac{1}{210}\right)=\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}+\frac{2}{195}\)
Hình như bn làm sai ấy. Chứ bài này sách của mk giải đc mà.
A=\(\frac{1}{42}+\frac{1}{56}+..............+\frac{1}{210}\)
=\(\frac{1}{6.7}+\frac{1}{7.8}+............+\frac{1}{14.15}\)
=\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+............+\frac{1}{14}-\frac{1}{15}\)
=\(\frac{1}{6}-\frac{1}{15}\)
=\(\frac{1}{10}\)
Đầu tiên: Rút gọn từng số hạng của A
Tiếp theo: Nhóm lại
Cuối cùng: Ra đáp án
A = 16/9 + 11/9 +9/11 +17/33 +11/39 + 2/13
A= (16/9+11/9) + (9/11+17/33) + (11/39+2/13)
A=3+4/3+17/39=62/13