Question

Tính.

a) \(\dfrac{2}{5}+\dfrac{3}{5}\)               \(\dfrac{1}{3}-\dfrac{2}{12}\)                  \(5+\dfrac{1}{4}\)                    \(1-\dfrac{1}{2}+\dfrac{1}{6}\)

b) \(\dfrac{2}{7}+\dfrac{1}{5}\)               \(\dfrac{2}{3}-\dfrac{5}{8}\)                    \(\dfrac{7}{8}+\dfrac{1}{6}\)                  \(1+\left(\dfrac{4}{7}-\dfrac{1}{4}\right)\) 

Answer 1

a) $\frac{2}{5} + \frac{3}{5} = \frac{5}{5} = 1$                                

$\frac{1}{3} - \frac{2}{{12}} = \frac{4}{{12}} - \frac{2}{{12}} = \frac{2}{{12}} = \frac{1}{6}$

$5 + \frac{1}{4} = \frac{{20}}{4} + \frac{1}{4} = \frac{{21}}{4}$                                     

  $1 - \frac{1}{2} + \frac{1}{6} = \frac{6}{6} - \frac{3}{6} + \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$

b) $\frac{2}{7} + \frac{1}{5} = \frac{{10}}{{35}} + \frac{7}{{35}} = \frac{{17}}{{35}}$                             

$\frac{2}{3} - \frac{5}{8} = \frac{{16}}{{24}} - \frac{{15}}{{24}} = \frac{1}{{24}}$

$\frac{7}{8} + \frac{1}{6} = \frac{{42}}{{48}} + \frac{8}{{48}} = \frac{{50}}{{48}} = \frac{{25}}{{24}}$                                   

$1 + \left( {\frac{4}{7} - \frac{1}{4}} \right) = 1 + \left( {\frac{{16}}{{28}} - \frac{7}{{28}}} \right) = 1 + \frac{9}{{28}} = \frac{{28}}{{28}} + \frac{9}{{28}} = \frac{{37}}{{28}}$