Question

Cho biểu thức

P=\(\frac{1}{2000.1999}\)-\(\frac{1}{1999.1998}\)-...-\(\frac{1}{3.2}\)-\(\frac{1}{2.1}\)

Hãy tính giá trị P+\(\frac{1997}{1999}\)

Answer 1

\(P=\frac{1}{2000.1999}-\frac{1}{1999.1998}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(\Rightarrow P=\frac{1}{2000.1999}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1998.1999}+\frac{1}{1999.2000}\right)\)

\(\Rightarrow P=\frac{1}{2000.1999}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1999}-\frac{1}{2000}\right)\)

\(\Rightarrow P=\frac{1}{1999.2000}-\left(1-\frac{1}{2000}\right)\)

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

\(\Rightarrow P=\frac{1}{1999}-1\)

\(\Rightarrow P=\frac{-1998}{1999}\)

\(\Rightarrow P+\frac{1997}{1999}=\frac{-1}{1999}\)

Vậy...

Answer 2

Thanks bn nha