Bài 1: Tính
A=\(\left(\dfrac{3}{7}-\dfrac{3}{17}+\dfrac{3}{37}\right):\left(\dfrac{5}{7}-\dfrac{5}{17}+\dfrac{5}{37}\right)+\dfrac{2}{5}\)
Bài 2: So sánh B với \(\dfrac{1}{2}\) biết:
B= \(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\)
Bài 1 :
\(A=\dfrac{\dfrac{3}{7}-\dfrac{3}{17}+\dfrac{3}{37}}{\dfrac{5}{7}-\dfrac{5}{17}+\dfrac{5}{37}}+\dfrac{2}{5}=\dfrac{3\left(\dfrac{1}{7}-\dfrac{1}{17}+\dfrac{1}{37}\right)}{5\left(\dfrac{1}{7}-\dfrac{1}{17}+\dfrac{1}{37}\right)}+\dfrac{2}{5}=\dfrac{3}{5}+\dfrac{2}{5}=\dfrac{5}{5}=1\)