Tính bằng cách thuận tiện: a) 12/15 + 35/52 +3/15 + 17/52 b) 8/9 * 12/5 x 9/8 x 7/6

a) \[\frac{{12}}{{15}}\]+ \[\frac{{35}}{{52}}\]+ \[\frac{3}{{15}}\] + \[\frac{{17}}{{52}}\]

= \[\left( {\frac{{12}}{{15}} + \frac{3}{{15}}} \right)\]+ \[\left( {\frac{{35}}{{52}} + \frac{{17}}{{52}}} \right)\]

= 1 + 1 = 2

b) \[\frac{8}{9}\]× \[\frac{{12}}{5}\]× \[\frac{9}{8}\]× \[\frac{7}{6}\]

= \[\left( {\frac{8}{9} \times \frac{9}{8}} \right)\]× \[\left( {\frac{{12}}{5} \times \frac{7}{6}} \right)\]

= 1 × \[\frac{{14}}{5}\]= \[\frac{{14}}{5}\]

c) \[\frac{1}{8}\]× \[\frac{3}{{14}}\]+ \[\frac{1}{8}\]× \[\frac{{11}}{{14}}\]

= \[\frac{1}{8}\]× \[\left( {\frac{3}{{14}} + \frac{{11}}{{14}}} \right)\]

= \[\frac{1}{8}\]× 1 = \[\frac{1}{8}\]

d) \[\frac{1}{6}\]× \[\frac{{13}}{{12}}\] – \[\frac{1}{6}\]× \[\frac{1}{{12}}\]

= \[\frac{1}{6}\]× \[\left( {\frac{{13}}{{12}} - \frac{1}{{12}}} \right)\]

= \[\frac{1}{6}\]× 1 = \[\frac{1}{6}\]