Hãy tính nhanh dãy số sau mà không cần làm phép tính :
a, 1+1-1+1-1+1-1+1-1+1-1+1-1+1-1
b, 1-1+1-1+1-1+1-1+1-1+1-1+1-1+1-1
c, 1+1-1+1-1+1-1+1-1+1-1+1-1+1-1+1
Đây là bài toán thật chứ không phạm quy nên xin OLM đừng xóa
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Bạn kiểm tra lại đề hộ. Nếu có phân số \(\frac{1}{4}\)thì chịu còn không có thì dễ.
Bài 1
\(\left(1-\dfrac{1}{99}\right)\times\left(1-\dfrac{1}{100}\right)\times...\times\left(1-\dfrac{1}{2006}\right)\)
\(=\dfrac{98}{99}\times\dfrac{99}{100}\times...\times\dfrac{2005}{2006}\)
\(=\dfrac{98}{2006}\)
\(=\dfrac{49}{1003}\)
Bài 2
\(\dfrac{111}{333}=\dfrac{111:111}{333:111}=\dfrac{1}{3}\)
\(\dfrac{2222}{4444}=\dfrac{2222:2222}{4444:2222}=\dfrac{1}{2}\)
Do \(3>2\Rightarrow\dfrac{1}{3}< \dfrac{1}{2}\)
Vậy \(\dfrac{111}{333}< \dfrac{2222}{4444}\)
Bài 1.
\(\left(1-\dfrac{1}{99}\right)\times\left(1-\dfrac{1}{100}\right)\times...\times\left(1-\dfrac{1}{2006}\right)\)
\(=\dfrac{98}{99}\times\dfrac{99}{100}\times...\times\dfrac{2005}{2006}\)
\(=\dfrac{98\times99\times...\times2005}{99\times100\times...2006}\)
\(=\dfrac{98}{2006}\)
\(=\dfrac{49}{1003}\)
Bài 2.
Có: \(\dfrac{111}{333}=\dfrac{111}{3\times111}=\dfrac{1}{3}\)
\(\dfrac{2222}{4444}=\dfrac{2222}{2\times2222}=\dfrac{1}{2}\)
Vì \(\dfrac{1}{3}< \dfrac{1}{2}\) nên \(\dfrac{111}{333}< \dfrac{2222}{4444}\)
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
`Answer:`
\(A=-\frac{1}{20}+-\frac{1}{30}+-\frac{1}{42}+-\frac{1}{56}+-\frac{1}{72}+-\frac{1}{90}\)
\(=-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}-\frac{1}{72}-\frac{1}{90}\)
\(=-\frac{1}{20}-\left(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=-\frac{1}{20}-\left(\frac{1}{5}-\frac{1}{10}\right)\)
\(=-\frac{1}{20}-\frac{1}{10}\)
\(=-\frac{3}{20}\)
A=\(\dfrac{-1}{20}+\dfrac{-1}{30}+\dfrac{-1}{42}+\dfrac{-1}{56}+\dfrac{-1}{72}+\dfrac{-1}{90}\)
A=\(-\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)\)
A=\(-\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
A=\(-\left(\dfrac{1}{4}-\dfrac{1}{10}\right)\)
A=\(-\dfrac{3}{20}\)
tách đc như bước 3 là nhờ công thức \(\dfrac{1}{n\left(n+1\right)}=\dfrac{1}{n}-\dfrac{1}{n+1}\) hoặc \(\dfrac{k}{n\left(n+k\right)}=\dfrac{1}{n}-\dfrac{1}{n+k}\) nhé
\(A=\dfrac{-1}{20}+\dfrac{-1}{30}+\dfrac{-1}{42}+\dfrac{-1}{56}+\dfrac{-1}{72}+\dfrac{-1}{90}\)
\(A=\dfrac{-1}{4\cdot5}+\dfrac{-1}{5\cdot6}+\dfrac{-1}{6\cdot7}+\dfrac{-1}{7\cdot8}+\dfrac{-1}{8\cdot9}+\dfrac{-1}{9\cdot10}\)
\(A=\dfrac{-1}{4}-\dfrac{-1}{5}+\dfrac{-1}{5}-\dfrac{-1}{6}+\dfrac{-1}{6}-\dfrac{-1}{7}+\dfrac{-1}{7}-\dfrac{-1}{8}+\dfrac{-1}{8}-\dfrac{-1}{9}+\dfrac{-1}{9}-\dfrac{-1}{10}\)
\(A=\dfrac{-1}{4}-\dfrac{-1}{10}\)
\(A=-\dfrac{3}{20}\)