c y hệt cn nít @@ 

-2011         

\(\left(-2\right)\left(-1\frac{1}{2}\right)...........\left(-1\frac{1}{2010}\right)=\frac{\left[\left(-2\right)\left(-3\right).....\left(-2010\right)\right].\left(-2011\right)}{\left(2.3.4.............2010\right)}=\frac{\left(-1\right)\left(-2011\right)}{1}=2011\)

\(\left(-2\right).\left(-1\frac{1}{2}\right).\left(-1\frac{1}{3}\right)....\left(-1\frac{1}{2009}\right).\left(-1\frac{1}{2010}\right)=\left(-2\right).\left(-\frac{3}{2}\right).\left(-\frac{4}{3}\right)....\left(-\frac{2010}{2009}\right).\left(-\frac{2011}{2010}\right)=\frac{\left(-2\right).\left(-3\right).\left(-4\right)....\left(-2010\right).\left(-2011\right)}{2.3.4....2009.2010}\)=2011

-2(-3/2) (-4/3) ...(-2010/2009) (-2011/2010) (co 2010 so hang )

=2011

\(\left(-2\right)\left(-1\frac{1}{2}\right)\left(-1\frac{1}{3}\right)...\left(-1\frac{1}{2009}\right)\left(-1\frac{1}{2010}\right)=\frac{-2}{1}.\frac{-3}{2}.\frac{-4}{3}....\frac{-2010}{2009}.\frac{-2011}{2010}=\frac{\left(2.3.4...2010\right).2011}{\left(1.2.3...2010\right)}=2011\)

(Từ 1 đến 2010 có 2010 số nên có 2010 dấu (-) => tích dương)

\(y=\left(-2\right)\left(-1\frac{1}{2}\right)\left(-1\frac{1}{3}\right).....\left(-1\frac{1}{2010}\right)\)

\(y=\left(-2\right)\left(\frac{-3}{2}\right)\left(\frac{-4}{3}\right)....\left(\frac{-2011}{2010}\right)\)

\(y=2011\)

y=-2011, violympic dung ko

mấy người muốn 8 thì nhắn tin mà 8, cứ đẩy bài người khác xuống
 

\(\left(-2\right).\left(-1\frac{1}{2}\right).\left(-1\frac{1}{3}\right)...\left(-1\frac{1}{2013}\right)\)

\(=\frac{-2}{1}.\frac{-3}{2}.\frac{-4}{3}...\frac{-2014}{2013}\)

\(=-\left(\frac{2}{1}.\frac{3}{2}.\frac{4}{3}...\frac{2014}{2013}\right)\)(vì có lẻ thừa số âm)

\(=-2014\)

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^^

\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{2012}-1\right)\left(\frac{1}{2013}-1\right)\)

\(=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}...\frac{-2011}{2012}.\frac{-2012}{2013}\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2011}{2012}.\frac{2012}{2013}\)(vì tích trên có 2012 thừa số, mỗi thừa số là âm nên kết quả là dương)

\(=\frac{1.2.3...2011.2012}{2.3.4...2012.2013}=\frac{1}{2013}\)