Cho ba điểm \(A,\,\,B,\,\,C\) thẳng hàng cùng, biết \(AB = 3{\rm{\;cm}},\,\,BC = 4{\rm{\;cm}},\,\,AC = 7{\rm{\;cm}}.\)

a) Điểm \(B\) nằm giữa hai điểm \(A\) và \(C.\)

1) a) \(\frac{2}{7} + \frac{5}{7} \cdot \left( {60\% - 0,25} \right) \cdot {\left( { - 2} \right)^2}\)

\( = \frac{2}{7} + \frac{5}{7} \cdot \left( {\frac{{60}}{{100}} - \frac{{25}}{{100}}} \right) \cdot 4\)

\( = \frac{2}{7} + \frac{5}{7} \cdot 4 \cdot \frac{{35}}{{100}}\)

\( = \frac{2}{7} + \frac{{20}}{7}.\frac{7}{{20}}\)

\[ = \frac{2}{7} + 1\]\[ = \frac{9}{7}\].

b) \(\frac{2}{5} \cdot \left( {\frac{{ - 5}}{{12}} + \frac{{ - 9}}{{13}}} \right) - \frac{2}{5} \cdot \left( {\frac{8}{{13}} - \frac{5}{{12}}} \right):2\)

\( = \frac{1}{5} \cdot 2 \cdot \left( {\frac{{ - 5}}{{12}} + \frac{{ - 9}}{{13}}} \right) - \frac{2}{5} \cdot \left( {\frac{8}{{13}} - \frac{5}{{12}}} \right) \cdot \frac{1}{2}\)

\( = \frac{1}{5} \cdot \left( {\frac{{ - 10}}{{12}} + \frac{{ - 18}}{{13}}} \right) - \frac{1}{5} \cdot \left( {\frac{8}{{13}} - \frac{5}{{12}}} \right)\)

\[ = \frac{1}{5} \cdot \left( {\frac{{ - 10}}{{12}} + \frac{{ - 18}}{{13}} - \frac{8}{{13}} + \frac{5}{{12}}} \right)\]

\[ = \frac{1}{5} \cdot \left[ {\left( {\frac{{ - 10}}{{12}} + \frac{5}{{12}}} \right) + \left( {\frac{{ - 18}}{{13}} - \frac{8}{{13}}} \right)} \right]\]

\[ = \frac{1}{5} \cdot \left( {\frac{{ - 5}}{{12}} + \frac{{ - 26}}{{13}}} \right) = \frac{1}{5} \cdot \left[ {\frac{{ - 5}}{{12}} + \left( { - 2} \right)} \right]\]

\[ = \frac{1}{5} \cdot \frac{{ - 29}}{{12}} = \frac{{ - 29}}{{60}}.\]