A=15.3^11+4.3^12 phần 3^14 B= 5.2^13.2^22-2^36 phần (3.2^17)^2 C= 2^19.3^9+15.2^18.3^8 phần 2^9.3^9.2^10+3^10-2^10 D=2^14.2^8 phần 3.2^15 .2^8 -5.2^2

Giải thích các bước giải:

Ta có:

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