Question

Tính nhanh:\(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\) 

b,\(\frac{45+16-17}{45x15+18}\)

HELP ME PLEASE!!

Answer 1

dễ mk nhìn là biết

Answer 2

Đặt A = \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)

3A = \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)

3A - A = (\(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)) - (\(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\))

2A = 1 - \(\frac{1}{729}\) = \(\frac{728}{729}\)

A = \(\frac{728}{729}:2=\frac{364}{729}\)