Question

tính giá trị biểu thức

1)\(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{1012}-1\right)\)

Answer 1

ta có : \(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{1012}-1\right)\)

\(=\left(\dfrac{-1}{2}\right)\left(\dfrac{-2}{3}\right)\left(\dfrac{-3}{4}\right)...\left(\dfrac{-1011}{1012}\right)\) có \(1011\) nhân tử

\(=\dfrac{\left(-1\right)\left(-2\right)\left(-3\right)...\left(-1011\right)}{2.3.4...1012}=\dfrac{-1.2.3...1011}{2.3.4...1012}=\dfrac{-1}{1012}\)