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\(\left(-2\right)\left(-1\frac{1}{2}\right)...........\left(-1\frac{1}{2010}\right)=\frac{\left[\left(-2\right)\left(-3\right).....\left(-2010\right)\right].\left(-2011\right)}{\left(2.3.4.............2010\right)}=\frac{\left(-1\right)\left(-2011\right)}{1}=2011\)
\(\left(-2\right).\left(-1\frac{1}{2}\right).\left(-1\frac{1}{3}\right)....\left(-1\frac{1}{2009}\right).\left(-1\frac{1}{2010}\right)=\left(-2\right).\left(-\frac{3}{2}\right).\left(-\frac{4}{3}\right)....\left(-\frac{2010}{2009}\right).\left(-\frac{2011}{2010}\right)=\frac{\left(-2\right).\left(-3\right).\left(-4\right)....\left(-2010\right).\left(-2011\right)}{2.3.4....2009.2010}\)=2011
-2(-3/2) (-4/3) ...(-2010/2009) (-2011/2010) (co 2010 so hang )
=2011
\(\left(-2\right)\left(-1\frac{1}{2}\right)\left(-1\frac{1}{3}\right).....\left(-1\frac{1}{2010}\right)\)
= \(-2.\left(\frac{-3}{2}\right).\left(\frac{-4}{3}\right).....\left(\frac{-2011}{2010}\right)\)
= 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
\(=\left(-2\right).\left(-\frac{3}{2}\right).\left(-\frac{4}{3}\right)....\left(-\frac{2010}{2009}\right).\left(-\frac{2011}{2010}\right)=\frac{\left(-2\right).\left(-3\right).\left(-4\right)....\left(-2010\right).\left(-2011\right)}{2.3.4....2009.2010}=2011\)