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Ta viết lại biểu thức A như sau:
\(A=-\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{11.12}\right)\)
\(A=-\left(\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}+...+\dfrac{12-11}{11.12}\right)\)
\(A=-\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{12}\right)\)
\(A=-\left(\dfrac{1}{4}-\dfrac{1}{12}\right)\)
\(A=-\dfrac{1}{6}\)
Sửa đề: A=-1/20+(-1/30)+(-1/42)+(-1/56)+(-1/72)+(-1/90)
=-(1/20+1/30+...+1/90)
=-(1/4-1/5+1/5-1/6+...+1/9-1/10)
=-1/4+1/10
=-5/20+2/20=-3/20
A=1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10
A=1/2-1/3+1/3-1/4+1/4-1/5+...+1/9-1/10
A=1/2-1/10
A=2/5
A=-1/20+-1/30+-1/42+-1/56+-1/72+-1/90
Tìm Am,
Câu hỏi tương tự Đọc thêmToán lớp 6\(A=-\left(\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{9.10}\right)\)
\(=-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-....-\frac{1}{10}\right)=-\left(\frac{1}{4}-\frac{1}{10}\right)=-\frac{3}{20}\)
\(\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
= \(\frac{-1}{4.5}+\frac{-1}{5.6}+\frac{-1}{6.7}+\frac{-1}{7.8}+\frac{-1}{8.9}+\frac{-1}{9.10}\)
= \(\left(-1\right).\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
= \(\left(-1\right).\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}+\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
= \(\left(-1\right).\left(\frac{1}{4}-\frac{1}{10}\right)\)
= \(\left(-1\right).\frac{3}{20}\)
= \(\frac{-3}{20}\)
6323121356214488888888888888888888888888888888888888888888888888888888888888888888
\(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
\(=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{4}-\frac{1}{10}=\frac{5}{20}-\frac{2}{20}=\frac{3}{20}\)