Question

[Trịnh Quang]

Tính giá trị biểu thức A=3^2/1.4+3^2/4.7+…+3^2/196.199

Answer 1

[Doãn Đức Duy]

A = \(\dfrac{3^2}{1.4}+\dfrac{3^2}{4.7}+...+\dfrac{3^2}{196.199}\)

A = \(\dfrac{3.3}{1.4}+\dfrac{3.3}{4.7}+...+\dfrac{3.3}{196.199}\)

A = \(3.\dfrac{3}{1.4}+3.\dfrac{3}{4.7}+...+3.\dfrac{3}{196.199}\)

A = \(3\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{196.199}\right)\)

A = \(3\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{196}-\dfrac{1}{199}\right)\)

A = \(3\left(1-\dfrac{1}{199}\right)\) = \(3.\dfrac{198}{199}\) = \(\dfrac{594}{199}\)