Tính bằng cách thuận tiện:

a) \(\frac{2}{9} + \frac{1}{3} + \frac{4}{9}\)

= \((\frac{2}{9} + \frac{4}{9}) + \frac{1}{3}\)

= \(\frac{6}{9} + \frac{1}{3}\)

= 1

b) \(\frac{5}{{12}} + \frac{3}{8} + \frac{7}{{12}} + \frac{1}{8}\)

= \((\frac{5}{{12}} + \frac{7}{{12}}) + (\frac{3}{8} + \frac{1}{8})\)

= \(1 + \frac{1}{2}\)

= \(\frac{3}{2}\)