Cho \( A = \frac{2}{5} + 0,(54) - 7,(2);B = \frac{5}{{11}}.\frac{{44}}{{20}} - \left( { - \frac{7}{{10}}} \right).2,(5) + \frac{{11}}{9}\) So sánh A và B
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Lời giải:
Báo saiTa có:
\(\begin{array}{l} 0,\left( {54} \right) = \frac{{54}}{{99}} = \frac{6}{{11}};\\ 7,(2) = 7 + 0,(2) = 7 + \frac{2}{9} = \frac{{65}}{9};\\ 2,(5) = 2 + 0,(5) = 2 + \frac{5}{9} = \frac{{23}}{9}\\ \to \begin{array}{*{20}{l}} {A = \frac{2}{5} + 0,\left( {54} \right) - 7,\left( 2 \right)}\\ { = \frac{2}{5} + \frac{6}{{11}} - \frac{{65}}{9}}\\ { = \frac{{198}}{{495}} + \frac{{270}}{{495}} - \frac{{3575}}{{495}}}\\ { = \frac{{ - 3107}}{{495}}} \end{array}\\ \begin{array}{*{20}{l}} {B = \frac{5}{{11}}.\frac{{44}}{{20}} - \left( { - \frac{7}{{10}}} \right).2,\left( 5 \right) + \frac{{11}}{9}}\\ { = \frac{5}{{11}}.\frac{{11}}{4} - \left( {\frac{{ - 7}}{{10}}} \right).\frac{{23}}{9} + \frac{{11}}{9}}\\ { = \frac{5}{4} - \left( {\frac{{ - 161}}{{90}}} \right) + \frac{{11}}{9}}\\ { = \frac{{225}}{{180}} + \frac{{322}}{{180}} + \frac{{220}}{{180}}}\\ { = \frac{{767}}{{180}}} \end{array} \end{array}\)
Nhận thấy
\(\begin{array}{l} A = \frac{{ - 3107}}{{495}} < 0\\ B = \frac{{767}}{{180}} > 0\\ \to B > A \end{array}\)
