Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
a,\\
{\left( { - x} \right)^9}:{\left( { - x} \right)^5} = {\left( { - x} \right)^{9 - 5}} = {\left( { - x} \right)^4} = {x^4}\\
b,\\
{25^9}:{5^{12}} = {\left( {{5^2}} \right)^9}:{5^{12}} = {5^{18}}:{5^{12}} = {5^{18 - 12}} = {5^6}\\
e,\\
\left( { - \frac{{12}}{{25}}{x^4}{y^3}{z^5}} \right):\frac{4}{5}{x^4}y{z^2}\\
= \left( { - \frac{{12}}{{25}}:\frac{4}{5}} \right).\left( {{x^4}:{x^4}} \right).\left( {{y^3}:y} \right).\left( {{z^5}:{z^2}} \right)\\
= - \frac{3}{5}.1.{y^2}.{z^3}\\
= - \frac{3}{5}{y^2}{z^3}\\
h,\\
\frac{1}{3}{\left( {x + y} \right)^3}:\left( { - \frac{4}{5}} \right){\left( {x + y} \right)^2}\\
= \left( {\frac{1}{3}:\left( { - \frac{4}{5}} \right)} \right).\left[ {{{\left( {x + y} \right)}^3}:{{\left( {x + y} \right)}^2}} \right]\\
= - \frac{5}{{12}}.\left( {x + y} \right)\\
2,\\
a,\\
\left( {7{x^2}{y^3} + 2{x^2}{y^3}} \right):\left( { - 3{x^2}} \right)\\
= \left[ {{x^2}\left( {7{y^3} + 2{y^3}} \right)} \right]:\left( { - 3{x^2}} \right)\\
= \left( {{x^2}.9{y^3}} \right):\left( { - 3{x^2}} \right)\\
= - 3{y^3}\\
c,\\
\left( {8{x^4} - {x^3} + 3{x^2}} \right):2{x^2}\\
= \left[ {{x^2}.\left( {8{x^2} - x + 3} \right)} \right]:2{x^2}\\
= \left( {8{x^2} - x + 3} \right):2\\
= 4{x^2} - \frac{1}{2}x + \frac{3}{2}\\
g,\\
\left( { - 18{x^3}{y^5} + 12{x^2}{y^2} - 6x{y^3}} \right):6xy\\
= \left[ {6xy\left( { - 3{x^2}{y^4} + 2xy - {y^2}} \right)} \right]:6xy\\
= - 3{x^2}{y^4} + 2xy - {y^2}\\
h,\\
\left( {2{x^2}{y^3} - 3{x^2}{y^2} - {x^2}y} \right):\left( { - \frac{1}{3}{x^2}y} \right)\\
= \left[ {{x^2}y\left( {2{y^2} - 3y - 1} \right)} \right]:\left( { - \frac{1}{3}{x^2}y} \right)\\
= \left( {2{y^2} - 3y + 1} \right):\left( { - \frac{1}{3}} \right)\\
= - 6{y^2} + 9y - 3
\end{array}\)
