So sánh \[A = \frac{{{2^5}.7 + {2^5}}}{{{2^5}{{.5}^2} - {2^5}.3}}\] và \[\frac{{{3^4}.5 - {3^6}}}{{{3^4}.13 + {3^4}}}\] với 1

Trả lời:

\[\frac{{{2^5}.7 + {2^5}}}{{{2^5}{{.5}^2} - {2^5}.3}} = \frac{{{2^5}.\left( {7 + 1} \right)}}{{{2^5}.\left( {{5^2} - 3} \right)}} = \frac{{{2^5}.\left( {7 + 1} \right)}}{{{2^5}.\left( {25 - 3} \right)}} = \frac{{{2^5}.8}}{{{2^5}.22}} = \frac{8} = \frac{4}\]

\[\frac{{{3^4}.5 - {3^6}}}{{{3^4}.13 + {3^4}}} = \frac{{{3^4}.\left( {5 - {3^2}} \right)}}{{{3^4}.\left( {13 + 1} \right)}} = \frac{{{3^4}.\left( {5 - 9} \right)}}{{{3^4}.\left( {13 + 1} \right)}} = \frac{{{3^4}.\left( { - 4} \right)}}{{{3^4}.14}} = \frac{{ - 4}} = \frac{{ - 2}}{7}\]

MSC = 77

\[\frac{4} = \frac = \frac;\frac{{ - 2}}{7} = \frac{{ - 2.11}} = \frac{{ - 22}}\]

Do đó: \[\frac{{ - 22}} < \frac < 1\] hay B < A < 1

Đáp án cần chọn là: D