1.

a) \(11\frac{3}{{13}} - \left( {2\frac{4}{7} + 5\frac{3}{{13}}} \right) = \left( {11 + \frac{3}{{13}}} \right) - \left[ {\left( {2 + \frac{4}{7}} \right) + \left( {5 + \frac{3}{{13}}} \right)} \right]\)

\( = 11 + \frac{3}{{13}} - \left[ {7 + \frac{4}{7} + \frac{3}{{13}}} \right] = 11 + \frac{3}{{13}} - 7 - \frac{4}{7} - \frac{3}{{13}} = 4 - \frac{4}{7} = \frac{{24}}{7}\);

b) \[\frac{{ - 5}}{7} \cdot \frac{2}{{11}} + \frac{{ - 5}}{7} \cdot \frac{9}{{11}} + \frac{5}{7} = \frac{{ - 5}}{7}\left( {\frac{2}{{11}} + \frac{9}{{11}}} \right) + \frac{5}{7} = \frac{{ - 5}}{7} \cdot 1 + \frac{5}{7} = 0\].

2.

a) \[\frac{3}{7} + x = \frac{4}{5}\]

\[x = \frac{4}{5} - \frac{3}{7}\]

\[x = \frac{{28}}{{35}} - \frac{{15}}{{35}}\]

\[x = \frac{{13}}{{35}}\]

Vậy \[x = \frac{{13}}{{35}}\].

b) \[\left( {3x - 1} \right)\left( {\frac{{ - 1}}{2}x + 5} \right) = 0\]

• TH1: \[3x--1 = 0\]

\[3x = 1\]

\[x = \frac{1}{3}\].

• TH2: \[\frac{{ - 1}}{2}x + 5 = 0\]

\[\frac{{ - 1}}{2}x = - 5\]

\[x = - 5:\left( {\frac{{ - 1}}{2}} \right)\]

\[x = 10\].

Vậy \[x \in \left\{ {\frac{1}{3};\,\,10} \right\}\].

Số đối của phân số 4/5 là