Question

[nguyen thu trang]

Tính\(\dfrac{0,5+0,\left(3\right)-0,1\left(6\right)}{2.5+1,\left(6\right)-0,8\left(3\right)}\)

Answer 1

[@Nk>↑@]

Đặt giá trị biểu thức trên là A, ta có:

A=\(\dfrac{0,5+0,\left(3\right)-0,1\left(6\right)}{2,5+1,\left(6\right)-0,8\left(3\right)}\)

A=\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{6}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{6}}\)

A=\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{6}}{5.\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\right)}\)

A=\(\dfrac{1}{5}\)