Question

 

3	+	3	+ … +	3	=
6 x 8		8 x 10		158 x 160

Answer 1

Đặt A = \(\dfrac{3}{6\cdot8}+\dfrac{3}{8\cdot10}+...+\dfrac{3}{158\cdot160}\)

A = \(\dfrac{3}{2}\cdot\left(\dfrac{2}{6\cdot8}+\dfrac{2}{8\cdot10}+...+\dfrac{2}{158\cdot160}\right)\)

A = \(\dfrac{3}{2}\cdot\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}+...+\dfrac{1}{158}-\dfrac{1}{160}\right)\)

A = \(\dfrac{3}{2}\cdot\left(\dfrac{1}{6}-\dfrac{1}{160}\right)\)

A = \(\dfrac{3}{2}\cdot\dfrac{77}{480}\)

A = \(\dfrac{77}{320}\)

Vậy: A = \(\dfrac{77}{320}\)