Cho $A = \dfrac{2}{5} + 0,\left( {54} \right) - 7,\left( 2 \right)$ và  $B = \dfrac{5}{{11}}.\dfrac{{44}}{{20}} - \left( { - \dfrac{7}{{10}}} \right).2,\left( 5 \right) + \dfrac{{11}}{9}$. So sánh $A$ và $B$.

Ta có: $0,\left( {54} \right) = \dfrac{{54}}{{99}} = \dfrac{6}{{11}};$ $7,(2) = 7 + 0,(2) = 7 + \dfrac{2}{9} = \dfrac{{65}}{9};$  $2,(5) = 2 + 0,(5) = 2 + \dfrac{5}{9} = \dfrac{{23}}{9}$

$\begin{array}{l}A = \dfrac{2}{5} + 0,\left( {54} \right) - 7,\left( 2 \right)\\ = \dfrac{2}{5} + \dfrac{6}{{11}} - \dfrac{{65}}{9}\\ = \dfrac{{198}}{{495}} + \dfrac{{270}}{{495}} - \dfrac{{3575}}{{495}}\\ = \dfrac{{ - 3107}}{{495}}\end{array}$

$\begin{array}{l}B = \dfrac{5}{{11}}.\dfrac{{44}}{{20}} - \left( { - \dfrac{7}{{10}}} \right).2,\left( 5 \right) + \dfrac{{11}}{9}\\ = \dfrac{5}{{11}}.\dfrac{{11}}{4} - \left( {\dfrac{{ - 7}}{{10}}} \right).\dfrac{{23}}{9} + \dfrac{{11}}{9}\\ = \dfrac{5}{4} - \left( {\dfrac{{ - 161}}{{90}}} \right) + \dfrac{{11}}{9}\\ = \dfrac{{225}}{{180}} + \dfrac{{322}}{{180}} + \dfrac{{220}}{{180}}\\ = \dfrac{{767}}{{180}}\end{array}$

Nhận thấy $A = \dfrac{{ - 3107}}{{495}} < 0$ và $B = \dfrac{{767}}{{180}} > 0$ nên $B > A.$