1,tính a,45^10*5^18/75^15 b,2^10*3^8+6^8*20/4^5*9^4-2*6^9

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
\dfrac{{{{45}^{10}}{{.5}^{18}}}}{{{{75}^{15}}}} = \dfrac{{{{\left( {9.5} \right)}^{10}}{{.5}^{18}}}}{{{{\left( {3.25} \right)}^{15}}}} = \dfrac{{{9^{10}}{{.5}^{10}}{{.5}^{18}}}}{{{3^{15}}{{.25}^{15}}}} = \dfrac{{{{\left( {{3^2}} \right)}^{10}}{{.5}^{28}}}}{{{3^{15}}.{{\left( {{5^2}} \right)}^{15}}}} = \dfrac{{{3^{20}}{{.5}^{28}}}}{{{3^{15}}{{.5}^{30}}}} = \dfrac{{{3^5}}}{{{5^2}}}\\
b,\\
\dfrac{{{2^{10}}{{.3}^8} + {6^8}.20}}{{{4^5}{{.9}^4} – {{2.6}^9}}} = \dfrac{{{2^{10}}{{.3}^8} + {{\left( {2.3} \right)}^8}{{.2}^2}.5}}{{{{\left( {{2^2}} \right)}^5}.{{\left( {{3^2}} \right)}^4} – 2.{{\left( {2.3} \right)}^9}}}\\
 = \dfrac{{{2^{10}}{{.3}^8} + {2^8}{{.3}^8}{{.2}^2}.5}}{{{2^{10}}{{.3}^8} – {{2.2}^9}{{.3}^9}}}\\
 = \dfrac{{{2^{10}}{{.3}^8} + {2^{10}}{{.3}^8}.5}}{{{2^{10}}{{.3}^8} – {2^{10}}{{.3}^9}}}\\
 = \dfrac{{{2^{10}}{{.3}^8}.\left( {1 + 5} \right)}}{{{2^{10}}{{.3}^8}.\left( {1 – 3} \right)}}\\
 = \dfrac{{1 + 5}}{{1 – 3}} = \dfrac{6}{{ – 2}} =  – 3
\end{array}\)