Tính giá trị biểu thức: \(M = a \cdot \frac{5}{9} + a \cdot \frac{4}{9} + 3\frac{5}{7}\) với \(a = \frac{{ - 5}}{7}\), ta được:

\(A = 92 - \frac{1}{9} - \frac{2}{{10}} - \frac{3}{{11}} - ... - \frac{{92}}{{100}}\)

\( = \left( {1 - \frac{1}{9}} \right) + \left( {1 - \frac{2}{{10}}} \right) + \left( {1 - \frac{3}{{11}}} \right) + ... + \left( {1 - \frac{{92}}{{100}}} \right)\) (có 92 số hạng)

\( = \frac{8}{9} + \frac{8}{{10}} + \frac{8}{{11}} + ... + \frac{8}{{100}}\)

\( = 8.\left( {\frac{1}{9} + \frac{1}{{10}} + \frac{1}{{11}} + ... + \frac{1}{{100}}} \right)\)

\(\begin{array}{l}B = \frac{1}{{45}} + \frac{1}{{50}} + \frac{1}{{55}} + ... + \frac{1}{{500}}\\ = \frac{1}{{5.9}} + \frac{1}{{5.10}} + \frac{1}{{5.11}} + ... + \frac{1}{{5.100}}\\ = \frac{1}{5}.\frac{1}{9} + \frac{1}{5}.\frac{1}{{10}} + \frac{1}{5}.\frac{1}{{11}} + ... + \frac{1}{5}.\frac{1}{{100}}\\ = \frac{1}{5}.\left( {\frac{1}{9} + \frac{1}{{10}} + \frac{1}{{11}} + ... + \frac{1}{{100}}} \right)\end{array}\)

\(\frac{A}{B} = \frac{{8.\left( {\frac{1}{9} + \frac{1}{{10}} + \frac{1}{{11}} + ... + \frac{1}{{100}}} \right)}}{{\frac{1}{5}.\left( {\frac{1}{9} + \frac{1}{{10}} + \frac{1}{{11}} + ... + \frac{1}{{100}}} \right)}} = \frac{8}{{\frac{1}{5}}} = 8.5 = 40\)

a) \(A = 26,3 - 7,63 + 13,7 + 200,4 - 12,37\).

\[A = \left( {26,3 + 13,7} \right) + \left( { - 7,63 - 12,37} \right) + 200,4\]

\[A = 40 + \left( { - 20} \right) + 200,4\]

\[A = 20 + 200,4\]

\[A = 220,4\].

b) \(B = \frac{{13}}{5}.\frac{{ - 3}}{7} + 2,6.\frac{{ - 4}}{7} + 2\frac{2}{5}\).

\(B = \frac{{13}}{5}.\frac{{ - 3}}{7} + \frac{{13}}{5}.\frac{{ - 4}}{7} + 2\frac{2}{5}\)

\(B = \frac{{13}}{5}.\left( {\frac{{ - 3}}{7} + \frac{{ - 4}}{7}} \right) + 2\frac{2}{5}\)

\(B = \frac{{13}}{5}.\left( { - 1} \right) + \frac{{12}}{5}\)

\(B = \frac{{ - 13}}{5} + \frac{{12}}{5}\)

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

c) \(C = \left( { - 9} \right).3 + 3.{\left( { - 2} \right)^3} - 49\).

\(C = - 27 + \left( { - 24} \right) - 49\)

\(C = - 100\).