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\(\left(-1\frac{1}{3}\right).\left(-1\frac{1}{4}\right).\left(-1\frac{1}{5}\right)....\left(-1\frac{1}{2012}\right)=\frac{-2}{3}.\frac{-3}{4}.....\frac{-2011}{2012}=\frac{2}{2012}=\frac{1}{1006}\)
hình như có chỗ nhầm:
\(=\frac{-4}{3}\cdot\frac{-5}{4}\cdot\frac{-6}{5}\cdot...\cdot\frac{-2013}{2012}\)
Ta áp dụng công thức: \(a-b=\left[-\left(b-a\right)\right]\)
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{2012}-1\right)\left(\frac{1}{2013}-1\right)\)
\(=-\left[\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2012}\right)\left(1-\frac{1}{2013}\right)\right]\)
\(=-\left(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2011}{2012}.\frac{2012}{2013}\right)\)
\(=-\frac{1.2.3...2011.2012}{2.3.4....2012.2013}\)
\(=-\frac{1}{2013}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{2012}{2013}\)
Liệt tử thừa với mẫu thừa:
\(=\frac{1}{2013}\)
Chúc em học tốt^^
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{2012}-1\right)\left(\frac{1}{2013}-1\right)\)
\(=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}...\frac{-2011}{2012}.\frac{-2012}{2013}\)
\(=\frac{\left(-1\right).\left(-2\right).\left(-3\right)...\left(-2011\right).\left(-2012\right)}{2.3.4....2013}\)
\(=\frac{1.2.3...2011.2012}{2.3.4.5...2013}\) ( vì các số hạng ở trên tử chẵn )
\(=\frac{1}{2013}\)
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{9}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{8}{9}\)
\(A=\frac{1}{9}\)
\(\Rightarrow\)A= \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}.\frac{7}{8}\frac{8}{9}\)
\(\Rightarrow\)A=\(\frac{1.2.3.4.5.6.7.8}{2.3.4.5.6.7.8.9}\)
\(\Rightarrow\)A=\(\frac{1}{9}\)
HỌC TỐT!!!
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:
\(\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+...+2012}\right)\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{4.7}{5.6}....\frac{2011.2014}{2012.2013}\)
\(=\frac{1}{3}.\frac{2014}{2012}=\frac{1007}{3018}\)
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{2012}-1\right)\left(\frac{1}{2013}-1\right)\)
\(=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}...\frac{-2011}{2012}.\frac{-2012}{2013}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2011}{2012}.\frac{2012}{2013}\)(vì có 2012 thừa số âm nên kết quả là dương)
\(=\frac{1}{2013}\)
\(\left(-1\frac{1}{3}\right).\left(-1\frac{1}{4}\right).\left(-1\frac{1}{5}\right)...\left(-1\frac{1}{2012}\right)\)
= \(-\frac{2}{3}.\frac{-3}{4}.\frac{-4}{5}...\frac{-2011}{2012}\)
= \(\frac{\left(-2\right).\left(-3\right).\left(-4\right)...\left(-2011\right)}{3.4.5...2012}\)
= \(\frac{2.3.4....2011}{3.4.5..2012}\)
= \(\frac{2}{2012}\)
= \(\frac{1}{1006}\)