Question

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\) tính

Answer 1

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+........+\frac{1}{99\cdot100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+........+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}=\frac{99}{100}\)

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Answer 2

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

Answer 3

A=\(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\)

A=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)

A=1-1/100

A=99/100

Answer 4

=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+......+\(\frac{1}{99}\)-\(\frac{1}{100}\)

=1-\(\frac{1}{100}\)

=\(\frac{99}{100}\)

Answer 5

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{100}{100}-\frac{1}{100}\)

\(=\frac{99}{100}\)