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Đáp án: $\begin{array}{l}
a)8{x^4}{y^2}{z^2} + 16{x^3}z - 2yz\\
b){x^2} - x - 2
\end{array}$

Giải thích các bước giải:

$\begin{array}{l}
a)\left( {4{x^5}{y^4}{z^3} + 8{x^4}{y^2}{z^2} - x{y^3}{z^2}} \right):\dfrac{1}{2}x{y^2}z\\
 = \left( {4{x^5}{y^4}{z^3} + 8{x^4}{y^2}{z^2} - x{y^3}{z^2}} \right).\dfrac{2}{{x{y^2}z}}\\
 = 4{x^5}{y^4}{z^3}.\dfrac{2}{{x{y^2}z}} + 8{x^4}{y^2}{z^2}.\dfrac{2}{{x{y^2}z}} - x{y^3}{z^2}.\dfrac{2}{{x{y^2}z}}\\
 = 8{x^4}{y^2}{z^2} + 16{x^3}z - 2yz\\
b)\left( {{x^4} - {x^2} - {x^3} - 2 - x} \right):\left( {{x^2} + 1} \right)\\
 = \left( {{x^4} + {x^2} - {x^3} - x - 2{x^2} - 2} \right):\left( {{x^2} + 1} \right)\\
 = \left[ {{x^2}\left( {{x^2} + 1} \right) - x\left( {{x^2} + 1} \right) - 2\left( {{x^2} + 1} \right)} \right]:\left( {{x^2} + 1} \right)\\
 = \left( {{x^2} + 1} \right)\left( {{x^2} - x - 2} \right):\left( {{x^2} + 1} \right)\\
 = {x^2} - x - 2
\end{array}$