Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = \dfrac{{{{15.3}^{11}} + {{4.3}^{12}}}}{{{3^{14}}}} = \dfrac{{{{5.3.3}^{11}} + {{4.3}^{12}}}}{{{3^{14}}}}\\
= \dfrac{{{{5.3}^{12}} + {{4.3}^{12}}}}{{{3^{14}}}} = \dfrac{{{{9.3}^{12}}}}{{{3^{14}}}} = \dfrac{{{3^2}{{.3}^{12}}}}{{{3^{14}}}} = \dfrac{{{3^{14}}}}{{{3^{14}}}}\\
B = \dfrac{{{{5.2}^{13}}{{.2}^{22}} - {2^{36}}}}{{{{\left( {{{3.2}^{17}}} \right)}^2}}} = \dfrac{{{{5.2}^{13 + 22}} - {{2.2}^{35}}}}{{{3^2}.{{\left( {{2^{17}}} \right)}^2}}}\\
= \dfrac{{{{5.2}^{35}} - {{2.2}^{35}}}}{{{3^2}{{.2}^{34}}}} = \dfrac{{{{3.2}^{35}}}}{{{3^2}{{.2}^{34}}}} = \dfrac{{1.2}}{{3.1}} = \dfrac{2}{3}\\
D = \dfrac{{{2^{14}}{{.2}^8}}}{{{{3.2}^{15}}{{.2}^8} - {{5.2}^2}{{.2}^{20}}}} = \dfrac{{{2^{14 + 8}}}}{{{{3.2}^{15 + 8}} - {{5.2}^{2 + 20}}}}\\
= \dfrac{{{2^{22}}}}{{{{3.2}^{23}} - {{5.2}^{22}}}} = \dfrac{{{2^{22}}}}{{{{3.2.2}^{22}} - {{5.2}^{22}}}} = \dfrac{{{2^{22}}}}{{{{6.2}^{22}} - {{5.2}^{22}}}} = \dfrac{{{2^{22}}}}{{{2^{22}}}} = 1
\end{array}\)
