A. A < 1 < B
B. A = B = 1
C. A >1 >B
D. 1 >A >B
Giải bởi Vietjack
\[\frac{{{2^5}.7 + {2^5}}}{{{2^5}{{.5}^2} - {2^5}.3}} = \frac{{{2^5}.(7 + 1)}}{{{2^5}.({5^2} - 3)}} = \frac{{{2^5}.(7 + 1)}}{{{2^5}.(25 - 3)}} = \frac{{{2^5}.8}}{{{2^5}.22}} = \frac{8}{{22}} = \frac{4}{{11}}\]
\[\frac{{{3^4}.5 - {3^6}}}{{{3^4}.13 + {3^4}}} = \frac{{{3^4}.(5 - {3^2})}}{{{3^4}.(13 + 1)}} = \frac{{{3^4}.(5 - 9)}}{{{3^4}.14}} = \frac{{{3^4}.( - 4)}}{{{3^4}.14}} = \frac{{ - 4}}{{14}} = \frac{{ - 2}}{7}\]
MSC = 77
\[\frac{4}{{11}} = \frac{{4.7}}{{11.7}} = \frac{{28}}{{77}};\frac{{ - 2}}{7} = \frac{{ - 2.11}}{{7.11}} = \frac{{ - 22}}{{77}}\]
Do đó \[\frac{{ - 22}}{{77}} < \frac{{28}}{{77}} < 1\] hay B < A < 1
Đáp án cần chọn là: D
