A. A = 18
B. A= 9
C. A = 54
D. A = 6
C
Ta có:
\(A = \frac{{{{11.3}^{22}}{{.3}^7} - {9^{15}}}}{{{{({{2.3}^{13}})}^2}}} = \frac{{{{11.3}^{22 + 7}} - {{({3^2})}^{15}}}}{{{2^2}.{{({3^{13}})}^2}}} = \frac{{{{11.3}^{29}} - {3^{2.15}}}}{{{2^2}{{.3}^{13.}}^2}} = \frac{{{{11.3}^{29}} - {3^{30}}}}{{{2^2}{{.3}^{26}}}} = \frac{{{{11.3}^{29}} - {3^{29}}.3}}{{{2^2}{{.3}^{26}}}} = \frac{{{3^{29}}.8}}{{{{4.3}^{26}}}} = {2.3^{29 - 26}} = {2.3^2} = 54\)Vậy A = 54
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