Question

[Phạm Quỳnh Anh]

TÍNH NHANH;

A= \(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+........+ \(\frac{1}{19.20}\)

Answer 1

[Xyz OLM]

A = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{19.20}\)

   = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\)

   = \(1-\frac{1}{20}\)

   = \(\frac{19}{20}\)

Vậy A = \(\frac{19}{20}\)

Answer 2

[Trang Thị Anh :)]

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{19.20}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\)

\(A=\frac{1}{1}-\frac{1}{20}=\frac{19}{20}\)

Vậy A = 19/20