Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\dfrac{{{2^{10}}{{.6}^{15}} + {3^{17}}{{.15.4}^{13}}}}{{{2^{18}}{{.18}^7}{{.3}^3} + {3^{15}}{{.2}^{25}}}}\\
= \dfrac{{{2^{10}}.{{\left( {2.3} \right)}^{15}} + {3^{17}}.3.5.{{\left( {{2^2}} \right)}^{13}}}}{{{2^{18}}.{{\left( {{{2.3}^2}} \right)}^7}{{.3}^3} + {3^{15}}{{.2}^{25}}}}\\
= \dfrac{{{2^{10}}{{.2}^{15}}{{.3}^{15}} + {3^{18}}{{.5.2}^{26}}}}{{{2^{18}}{{.2}^7}{{.3}^{14}}{{.3}^3} + {3^{15}}{{.2}^{25}}}}\\
= \dfrac{{{2^{25}}{{.3}^{15}} + {{5.3}^{18}}{{.2}^{26}}}}{{{2^{25}}{{.3}^{17}} + {3^{15}}{{.2}^{25}}}}\\
= \dfrac{{{2^{25}}{{.3}^{15}}.\left( {1 + {{5.3}^3}.2} \right)}}{{{2^{25}}{{.3}^{15}}.\left( {{3^2} + 1} \right)}}\\
= \dfrac{{{2^{25}}{{.3}^{15}}.271}}{{{2^{25}}{{.3}^{15}}.10}}\\
= \dfrac{{271}}{{10}}
\end{array}\)
