Cho $A = \dfrac{4}{9} + 1,2(31) + 0,(13)$ và $B = 3\dfrac{1}{2}.\dfrac{1}{{49}} - \left[ {2,\left( 4 \right).2\dfrac{5}{{11}}} \right]:\left( { - \dfrac{{28}}{9}} \right)$. So sánh $A$ và $B$.

Ta có $1,2\left( {31} \right) = 1 + 0,2\left( {31} \right)$ $= 1 + \dfrac{{231 - 2}}{{990}} = \dfrac{{1219}}{{990}}$ và $0,\left( {13} \right) = \dfrac{{13}}{{99}}$ ;

Lại có $2,\left( 4 \right) = 2 + 0,\left( 4 \right) = 2 + \dfrac{4}{9} = \dfrac{{22}}{9}$

Nên $A = \dfrac{4}{9} + 1,2(31) + 0,(13)$

         $= \dfrac{4}{9} + \dfrac{{1219}}{{990}} + \dfrac{{13}}{{99}} = \dfrac{{440 + 1219 + 130}}{{990}} = \dfrac{{1789}}{{990}}$

Và $B = 3\dfrac{1}{2}.\dfrac{1}{{49}} - \left[ {2,\left( 4 \right).2\dfrac{5}{{11}}} \right]:\left( { - \dfrac{{28}}{9}} \right)$

$= \dfrac{7}{2}.\dfrac{1}{{49}} - \left( {\dfrac{{22}}{9}.\dfrac{{27}}{{11}}} \right):\left( { - \dfrac{{28}}{9}} \right)$

$= \dfrac{1}{{14}} + 6.\dfrac{9}{{28}} = \dfrac{1}{{14}} + \dfrac{27}{14} = 2$

Nhận thấy $A = \dfrac{{1789}}{{990}} < \dfrac{{1980}}{{990}} = 2$ nên $A < B.$