Đáp án:
d) \(2{x^2} + x + 1\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\dfrac{{{x^3} - 8}}{{{x^2} + 2x + 4}} = \dfrac{{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}}{{{x^2} + 2x + 4}}\\
= x - 2\\
b)\dfrac{{3{x^3}{y^2}}}{{{x^2}}} = 3x{y^2}\\
c)\dfrac{{{x^3} + 2{x^2} - 2x - 1}}{{{x^2} + 3x + 1}}\\
= \dfrac{{{x^3} - {x^2} + 3{x^2} - 3x + x - 1}}{{{x^2} + 3x + 1}}\\
= \dfrac{{{x^2}\left( {x - 1} \right) + 3x\left( {x - 1} \right) + \left( {x - 1} \right)}}{{{x^2} + 3x + 1}}\\
= \dfrac{{\left( {x - 1} \right)\left( {{x^2} + 3x + 1} \right)}}{{{x^2} + 3x + 1}} = x - 1\\
d)\dfrac{{2{x^4} - 5{x^2} + {x^3} - 3 - 3x}}{{{x^2} - 3}}\\
= \dfrac{{2{x^4} - 6{x^2} + {x^2} - 3 + x\left( {{x^2} - 3} \right)}}{{{x^2} - 3}}\\
= \dfrac{{2{x^2}\left( {{x^2} - 3} \right) + \left( {{x^2} - 3} \right) + x\left( {{x^2} - 3} \right)}}{{{x^2} - 3}}\\
= \dfrac{{\left( {{x^2} - 3} \right)\left( {2{x^2} + x + 1} \right)}}{{{x^2} - 3}}\\
= 2{x^2} + x + 1
\end{array}\)
