Đáp án:
$\begin{array}{l}
+ ){\left( {x + y} \right)^2} - 25 = {\left( {x + y} \right)^2} - {5^2} = \left( {x + y - 5} \right)\left( {x + y + 5} \right)\\
+ )3{x^2} - 6xy + 3{y^2} = 3\left( {{x^2} - 2xy + {y^2}} \right) = 3{\left( {x - y} \right)^2}\\
+ )\left( {{x^4} - 2{x^3} + 4{x^2} - 8x} \right):\left( {{x^2} + 4} \right)\\
= \left[ {{x^3}\left( {x - 2} \right) + 4x\left( {x - 2} \right)} \right]:\left( {{x^2} + 4} \right)\\
= \left( {x - 2} \right)\left( {{x^3} + 4x} \right):\left( {{x^2} + 4} \right)\\
= \left( {x - 2} \right).x.\left( {{x^2} + 4} \right):\left( {{x^2} + 4} \right)\\
= \left( {x - 2} \right).x\\
+ )\frac{{{x^4} + 2{x^3} + 10x - 25}}{{{x^2} + 5}}\\
= \frac{{\left( {{x^4} - 25} \right) + \left( {2{x^3} + 10x} \right)}}{{{x^2} + 5}}\\
= \frac{{\left( {{x^2} + 5} \right)\left( {{x^2} - 5} \right) + 2x\left( {{x^2} + 5} \right)}}{{{x^2} + 5}}\\
= \frac{{\left( {{x^2} + 5} \right)\left( {{x^2} - 5 + 2x} \right)}}{{{x^2} + 5}}\\
= {x^2} + 2x - 5
\end{array}$
