Question

\(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{2017}-1\right)\)

Answer 1

\(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{2017}-1\right)\\ =\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot\dfrac{-3}{4}\cdot...\cdot\dfrac{-2016}{2017}\\ =\dfrac{\left(-1\right)\cdot\left(-2\right)\cdot\left(-3\right)\cdot...\cdot\left(-2016\right)}{2\cdot3\cdot4\cdot...\cdot2017}\\=\dfrac{\left(-1\right)\cdot\left(-1\right)\cdot\left(-1\right)\cdot...\left(-1\right)}{2017}\left(\text{có 2016 thừa số -1}\right)\\ =\dfrac{1}{2017}\)