Mk ko biết nhưng mk chúc bn sớm tìm đc câu trả lời
b)
\(\left(\dfrac{1}{10}-1\right)\left(\dfrac{1}{11}-1\right)\left(\dfrac{1}{12}-1\right)...\left(\dfrac{1}{2012}-1\right)\\ =\dfrac{-9}{10}\cdot\dfrac{-10}{11}\cdot\dfrac{-11}{12}\cdot...\cdot\dfrac{-2011}{2012}\\ =\left(-1\right)\cdot\dfrac{9}{10}\cdot\left(-1\right)\cdot\dfrac{10}{11}\cdot\left(-1\right)\cdot\dfrac{11}{12}\cdot...\cdot\left(-1\right)\cdot\dfrac{2011}{2012}\\ =\left[\left(-1\right)\cdot\left(-1\right)\cdot...\cdot\left(-1\right)\right]\cdot\left(\dfrac{9}{10}\cdot\dfrac{10}{11}\cdot\dfrac{11}{12}\cdot...\cdot\dfrac{2011}{2012}\right)\\ =\left[\left(-1\right)\cdot\left(-1\right)\cdot...\cdot\left(-1\right)\right]\cdot\dfrac{9}{2012}\)
(Có tất cả 2003 thừa số -1)
\(=\left(-1\right)\cdot\dfrac{9}{2012}=\dfrac{-9}{2012}\)