Question

Tính giá trị biểu thức

A=\(\dfrac{\left(1+17\right)\cdot\left(1+\dfrac{17}{2}\right)\cdot\left(1+\dfrac{17}{3}\right)....\left(1+\dfrac{17}{19}\right)}{\left(1+19\right)\cdot\left(1+\dfrac{19}{2}\right)\cdot\left(1+\dfrac{19}{3}\right)....\left(1+\dfrac{19}{17}\right)}\)

Answer 1

\(A=\dfrac{\left(1+17\right).\left(1+\dfrac{17}{2}\right)..........\left(1+\dfrac{17}{19}\right)}{\left(1+19\right).\left(1+\dfrac{19}{2}\right)..........\left(1+\dfrac{19}{17}\right)}\)

\(=\dfrac{18.\dfrac{19}{2}.............\dfrac{36}{19}}{20.\dfrac{21}{2}..........\dfrac{36}{17}}\)

\(=\dfrac{18.19.20.......36}{1.2.3...19}:\dfrac{20.21.....36}{1.2.3...17}\)

\(=\dfrac{1.2.3......36}{1.2.....36}\)

\(=1\)