Question

cÂU 1: tÍNH GIÁ TRỊ CỦA BIỂU THỨC:

A=\(\dfrac{9}{1.2}\) + \(\dfrac{9}{2.3}\) + \(\dfrac{9}{3.4}\) + ...... + \(\dfrac{9}{98.99}\) + \(\dfrac{9}{99.100}\)

Answer 1

A = \(\dfrac{9}{1.2}\)+ \(\dfrac{9}{2.3}\)+\(\dfrac{9}{3.4}\)+......+\(\dfrac{99}{99.100}\)

A = 9( \(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+.......+\(\dfrac{1}{99.100}\))

A = 9( 1-\(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+........+\(\dfrac{1}{99}\)-\(\dfrac{1}{100}\))

A = 9 ( 1 - \(\dfrac{1}{100}\))

A = 9 . \(\dfrac{99}{100}\)

A = \(\dfrac{891}{100}\)

Answer 2

\(A=\dfrac{9}{1\cdot2}+\dfrac{9}{2\cdot3}+\dfrac{9}{3\cdot4}+...+\dfrac{9}{98\cdot99}+\dfrac{9}{99\cdot100}\)

\(=9\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\right)\)

\(=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=9\left(1-\dfrac{1}{100}\right)\)

\(=9\left(\dfrac{100}{100}-\dfrac{1}{100}\right)\)

\(=9\cdot\dfrac{99}{100}\)

\(=\dfrac{891}{100}\)