Cho hai đường thẳng \(ab\) và \(cd\) cắt nhau tại \(O\). Góc đối đỉnh với góc \(aOc\) là

a) \(\frac{{21}}{{47}} + \frac{1}{5} + \frac{{26}}{{47}} - \frac{6}{5}\)

\( = \left( {\frac{{21}}{{47}} + \frac{{26}}{{47}}} \right) + \left( {\frac{1}{5} - \frac{6}{5}} \right)\)

\( = \frac{{47}}{{47}} + \left( { - \frac{5}{5}} \right)\)

\( = 1 + \left( { - 1} \right)\)

\( = 0.\)

b) \(15.{\left( { - \frac{1}{5}} \right)^2} + \frac{1}{5} - 2.{\left( { - \frac{1}{2}} \right)^2}\)

\( = 15.\frac{1}{{25}} + \frac{1}{5} - 2.\frac{1}{4}\)

\( = \frac{3}{5} + \frac{1}{5} - \frac{1}{2}\)

\( = \frac{4}{5} - \frac{1}{2}\)

\( = \frac{8}{{10}} - \frac{5}{{10}}\)

\( = \frac{3}{{10}}\).

c) \(\frac{4}{5}.\frac{3}{7} - \frac{{ - 4}}{7}:\sqrt {\frac{{25}}{{16}}} - \left| { - {{100}^0}} \right|\)

\( = \frac{4}{5} \cdot \frac{3}{7} - \frac{{ - 4}}{7}:\sqrt {{{\left( {\frac{5}{4}} \right)}^2}} - \left| { - 1} \right|\)

\( = \frac{4}{5} \cdot \frac{3}{7} + \frac{4}{7}:\frac{5}{4} - 1\)

\( = \frac{4}{5} \cdot \frac{3}{7} + \frac{4}{7} \cdot \frac{4}{5} - 1\)

\( = \frac{4}{5} \cdot \left( {\frac{3}{7} + \frac{4}{7}} \right) - 1\)

\( = \frac{4}{5} \cdot \frac{7}{7} - 1 = \frac{4}{5} \cdot 1 - 1\)

\( = \frac{4}{5} - 1 = \frac{{ - 1}}{5}.\)