Question

[_ Yuki _ Dễ thương _]

Cho biểu thức \(P=\dfrac{1}{2000.1999}-\dfrac{1}{1999.1998}-\dfrac{1}{1998.1997}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

Kết quả phép tính \(P+\dfrac{1997}{1999}\) là ?

Answer 1

[lê thị uyên]

-1/2000

Answer 2

[Hoang Hung Quan]

\(P=\dfrac{1}{2000.1999}-\dfrac{1}{1999.1998}-\dfrac{1}{1998.1997}-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(\Rightarrow P=\dfrac{1}{1999.2000}-\dfrac{1}{1998.1999}-\dfrac{1}{1997.1998}-\dfrac{1}{2.3}-\dfrac{1}{1.2}\)

\(\Rightarrow P=\dfrac{1}{1999}-\dfrac{1}{2000}-\dfrac{1}{1998}+\dfrac{1}{1999}-\dfrac{1}{1997}+\dfrac{1}{1998}-...-1+\dfrac{1}{2}\)

\(\Rightarrow P=\dfrac{2}{1999}-\dfrac{1}{2000}-1\)

\(\Rightarrow P+\dfrac{1997}{1999}=\dfrac{2}{1999}+\dfrac{1997}{1999}-\dfrac{1}{2000}-1\)

\(\Rightarrow P+\dfrac{1997}{1999}=1-1-\dfrac{1}{2000}=\dfrac{-1}{1200}\)

Vậy \(P+\dfrac{1997}{1999}=\dfrac{-1}{2000}\)