Question

 1/99x98 - ........- 1/3x2 - 1/2x1

Answer 1

Đặt \(A=\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(-A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\)

\(-A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\)

\(-A=1-\frac{1}{99}\)

\(-A=\frac{98}{99}\)

\(A=\frac{-98}{99}\)

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Answer 2

Đặt A = \(\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

=> - A = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\)

- A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\)

- A = \(1-\frac{1}{99}\)

- A = \(\frac{98}{99}\)

=> A = \(-\frac{98}{99}\)

Vậy A = \(-\frac{98}{99}\)

Hok tốt

Answer 3

\(A=\frac{1}{99.98}-...-\frac{1}{2.1}\)

\(-A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\)

\(-A=\frac{1}{1}-\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\)

\(-A=1-\frac{1}{99}\)

\(-A=\frac{98}{99}\Leftrightarrow A=\frac{-98}{99}\)