Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
a,\\
\left( {24{x^5} - 9{x^3} + 18{x^2}} \right):3x\\
= \left[ {3x\left( {8{x^4} - 9{x^2} + 6x} \right)} \right]:3x\\
= 8{x^4} - 9{x^2} + 6x\\
b,\\
\left( { - 4{x^2} + {x^3} - 20 + 5x} \right):\left( {x - 4} \right)\\
= \left[ {\left( {{x^3} - 4{x^2}} \right) + \left( {5x - 20} \right)} \right]:\left( {x - 4} \right)\\
= \left[ {{x^2}\left( {x - 4} \right) + 5\left( {x - 4} \right)} \right]:\left( {x - 4} \right)\\
= \left[ {\left( {x - 4} \right)\left( {{x^2} + 5} \right)} \right]:\left( {x - 4} \right)\\
= {x^2} + 5\\
2,\\
\left( {{{7.3}^5} - {3^4} + {3^6}} \right):{3^4}\\
= \left[ {{3^4}.\left( {7.3 - 1 + {3^2}} \right)} \right]:{3^4}\\
= 7.3 - 1 + {3^2}\\
= 21 - 1 + 9\\
= 29\\
b,\\
\left( {{{16}^3} - {{64}^2}} \right):{8^3}\\
= \left[ {{{\left( {2.8} \right)}^3} - {{\left( {{8^2}} \right)}^2}} \right]:{8^3}\\
= \left[ {{2^3}{{.8}^3} - {8^4}} \right]:{8^3}\\
= \left( {{{8.8}^3} - {8^4}} \right):{8^3}\\
= 0:{8^3}\\
= 0
\end{array}\)
