Question

A = \(\frac{1}{1\text{ X}3}\) + \(\frac{1}{3\text{ X}5}\)+...........+ \(\frac{1}{99\text{ X}101}\)

Answer 1

\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)

\(2A=1-\frac{1}{101}\)

\(2A=\frac{100}{101}\)

\(A=\frac{100}{101}:2=\frac{50}{101}\)

Answer 2

Ta có : \(\frac{1}{1x3}+\frac{1}{3x5}+........+\frac{1}{99x100}\)

=> 2M = \(\left(\frac{2}{1x3}+\frac{2}{3x5}+......+\frac{2}{99x100}\right)\)

=> 2A = \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{99}-\frac{1}{100}\)

=> 2A = 1 - \(\frac{1}{100}\)

=> 2A = \(\frac{99}{100}\)

=> A = \(\frac{99}{100}:2\)

=> A = \(\frac{99}{200}\)

Answer 3

Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+..+\frac{1}{99.101}\)

\(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)

\(2A=1-\frac{1}{101}\)

\(A=\frac{100}{101}:2=\frac{50}{101}\)

Ủng hộ mk nha !! ^_^

Answer 4

\(Ta có : \(\frac{1}{1x3}+\frac{1}{3x5}+........+\frac{1}{99x100}\) => 2M = \(\left(\frac{2}{1x3}+\frac{2}{3x5}+......+\frac{2}{99x100}\right)\) => 2A = \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{99}-\frac{1}{100}\) => 2A = 1 - \(\frac{1}{100}\) => 2A = \(\frac{99}{100}\) => A = \(\frac{99}{100}:2\) => A = \(\frac{99}{200}\)\)