\(\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\left(1-\frac{1}{1+2+3+4}\right).....\left(1-\frac{1}{1+2+3+...+2012}\right)\)
Những câu hỏi liên quan
Rút gọn :
a/ \(A=\frac{\frac{1}{19}+\frac{2}{18}+\frac{3}{17}+...+\frac{19}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{20}}\)
b/ \(B=\frac{\left(1+\frac{2012}{1}\right)\left(1+\frac{2012}{2}\right)...\left(1+\frac{2012}{1000}\right)}{\left(1+\frac{1000}{1}\right)\left(1+\frac{1000}{2}\right)...\left(1+\frac{1000}{2012}\right)}\)
Tính: A= \(\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\left(1-\frac{1}{1+2+3+4}\right)...\left(1-\frac{1}{1+2+3+...+2012}\right)\)
Ta có: \(1+2+...+n=\frac{\left(n+1\right)n}{2}\)
\(\Rightarrow\frac{1}{1+2+...+n}=\frac{2}{n\left(n+1\right)}\)
\(1-\frac{1}{1+2+...+n}=1-\frac{2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}\)
\(=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
Vậy nên:
\(\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+...+2012}\right)\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{4.7}{5.6}....\frac{2011.2014}{2012.2013}\)
\(=\frac{1}{3}.\frac{2014}{2012}=\frac{1007}{3018}\)
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Kết qủa của phép tính: \(\left(-2\right).\left(-1\frac{1}{2}\right).\left(-1\frac{1}{3}\right).\left(-1\frac{1}{4}\right)...\left(-1\frac{1}{2012}\right).\left(-1\frac{1}{2013}\right)\)
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{2012}-1\right)\left(\frac{1}{2013}-1\right)\)
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{2012}-1\right)\left(\frac{1}{2013}-1\right)\)
\(=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}...\frac{-2011}{2012}.\frac{-2012}{2013}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2011}{2012}.\frac{2012}{2013}\)(vì có 2012 thừa số âm nên kết quả là dương)
\(=\frac{1}{2013}\)
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left(frac{-5}{12}+frac{7}{4}-frac{3}{8}right)-left[4frac{1}{2}-7frac{1}{3}right]-left(frac{1}{4}-frac{5}{2}right)left[2frac{1}{4}-5frac{3}{2}right]-left(frac{3}{10}-1right)-5frac{1}{2}+left(frac{1}{3}-frac{5}{6}right)frac{4}{7}-left(3frac{2}{5}-1frac{1}{2}right)-frac{5}{21}+left[3frac{1}{2}-4frac{2}{3}right]frac{1}{8}-1frac{3}{4}+left(frac{7}{8}-3frac{7}{2}+frac{3}{4}right)-left[frac{7}{4}-frac{5}{8}right]left(frac{3}{5}-2frac{1}{10}+frac{11}{20}right)-left[frac{-3}{4}+1frac{7}{2}right]left[-2fr...
Đọc tiếp
\(\left(\frac{-5}{12}+\frac{7}{4}-\frac{3}{8}\right)-\left[4\frac{1}{2}-7\frac{1}{3}\right]-\left(\frac{1}{4}-\frac{5}{2}\right)\)
\(\left[2\frac{1}{4}-5\frac{3}{2}\right]-\left(\frac{3}{10}-1\right)-5\frac{1}{2}+\left(\frac{1}{3}-\frac{5}{6}\right)\)
\(\frac{4}{7}-\left(3\frac{2}{5}-1\frac{1}{2}\right)-\frac{5}{21}+\left[3\frac{1}{2}-4\frac{2}{3}\right]\)
\(\frac{1}{8}-1\frac{3}{4}+\left(\frac{7}{8}-3\frac{7}{2}+\frac{3}{4}\right)-\left[\frac{7}{4}-\frac{5}{8}\right]\)
\(\left(\frac{3}{5}-2\frac{1}{10}+\frac{11}{20}\right)-\left[\frac{-3}{4}+1\frac{7}{2}\right]\)
\(\left[-2\frac{1}{5}-2\frac{2}{3}\right]-\left(\frac{1}{15}-5\frac{1}{2}\right)+\left[\frac{-1}{6}+\frac{1}{3}\right]\)
\(1\frac{1}{8}-\left(\frac{1}{15}-\frac{1}{2}+\frac{-1}{6}\right)+\left[\frac{5}{4}+\frac{3}{2}\right]\)
\(\frac{5}{6}-\left(1\frac{1}{3}-1\frac{1}{2}\right)+\left[\frac{5}{12}-\frac{3}{4}-\frac{1}{6}\right]\)
\(1\frac{1}{4}-\left(\frac{7}{12}-\frac{2}{3}-1\frac{3}{8}\right)+\left[\frac{5}{24}-2\frac{1}{2}\right]-\frac{1}{6}-\left[\frac{-3}{4}\right]\)
\(-2\frac{1}{5}+2\frac{3}{10}-\left(\frac{6}{20}-\left[\frac{2}{8}-1\frac{1}{2}\right]\right)+\left[\frac{7}{20}-1\frac{1}{4}\right]\)
\(-\left[1\frac{2}{3}-3\frac{1}{2}+\frac{1}{4}\right]+\left(\frac{2}{6}-\frac{5}{12}\right)-\left(\frac{1}{3}-\left[\frac{1}{4}-\frac{1}{3}\right]\right)\)
\(-\frac{4}{5}-\left(1\frac{1}{10}-\frac{7}{10}\right)+\left[\frac{3}{4}-1\frac{1}{5}\right]+1\frac{1}{2}\)
\(\frac{3}{21}-\frac{5}{14}+\left[1\frac{1}{3}-5\frac{1}{2}+\frac{5}{14}\right]-\left(\frac{1}{6}-\frac{3}{7}+\frac{1}{3}\right)\)
\(-1\frac{2}{5}+\left[1\frac{3}{10}-\frac{7}{20}-1\frac{1}{4}\right]-\left(\frac{1}{5}-\left[\frac{3}{4}-1\frac{1}{2}\right]\right)\)
\(2\frac{1}{3}-\left(\frac{1}{2}-2\frac{1}{6}+\frac{3}{4}\right)+\left[\frac{5}{12}-1\frac{1}{3}\right]-\frac{7}{8}+3\frac{1}{2}\)
\(2\frac{1}{4}-1\frac{3}{5}-\left(\frac{9}{20}-\frac{7}{10}\right)+\left[1\frac{3}{5}-2\frac{1}{2}\right]+\frac{3}{4}\)
\(\left[\frac{8}{3}-5\frac{1}{4}+\frac{1}{6}\right]-\frac{7}{4}+\frac{-5}{12}-\left(1-1\frac{1}{2}+\frac{1}{3}\right)\)
\(\left(\frac{1}{4}-\left[1\frac{1}{4}-\frac{7}{10}\right]+\frac{1}{2}\right)-2\frac{1}{5}-1\frac{3}{10}+\left[1-\frac{1}{2}\right]\)
TRÌNH BÀY GIÚP MÌNH NHA
tính
\(A=\left(1-\frac{1}{1+2}\right)\cdot\left(1-\frac{1}{1+2+3}\right)\cdot\left(1-\frac{1}{1+2+3+4}\right)\cdot...\cdot\left(1-\frac{1}{1+2+3+...+2012}\right)\)
Kết quả phép tính:\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{2012}-1\right)\left(\frac{1}{2013}-1\right)\)
Ta áp dụng công thức: \(a-b=\left[-\left(b-a\right)\right]\)
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{2012}-1\right)\left(\frac{1}{2013}-1\right)\)
\(=-\left[\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2012}\right)\left(1-\frac{1}{2013}\right)\right]\)
\(=-\left(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2011}{2012}.\frac{2012}{2013}\right)\)
\(=-\frac{1.2.3...2011.2012}{2.3.4....2012.2013}\)
\(=-\frac{1}{2013}\)
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\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{2012}{2013}\)
Liệt tử thừa với mẫu thừa:
\(=\frac{1}{2013}\)
Chúc em học tốt^^
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\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{2012}-1\right)\left(\frac{1}{2013}-1\right)\)
\(=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}...\frac{-2011}{2012}.\frac{-2012}{2013}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2011}{2012}.\frac{2012}{2013}\)(vì tích trên có 2012 thừa số, mỗi thừa số là âm nên kết quả là dương)
\(=\frac{1.2.3...2011.2012}{2.3.4...2012.2013}=\frac{1}{2013}\)
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tính
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{2012}-1\right)\left(\frac{1}{2013}-1\right)\)
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{2012}-1\right)\left(\frac{1}{2013}-1\right)\)
\(=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}...\frac{-2011}{2012}.\frac{-2012}{2013}\)
\(=\frac{\left(-1\right).\left(-2\right).\left(-3\right)...\left(-2011\right).\left(-2012\right)}{2.3.4....2013}\)
\(=\frac{1.2.3...2011.2012}{2.3.4.5...2013}\) ( vì các số hạng ở trên tử chẵn )
\(=\frac{1}{2013}\)
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Tính P=\(\left(1-\frac{1}{1+2}\right)\cdot\left(1-\frac{1}{1+2+3}\right)\cdot\left(1-\frac{1}{1+2+3+4}\right)\cdot...\cdot\left(1-\frac{1}{1+2+3+...+2012}\right)\)