Đáp án:
$\begin{array}{l}
B6)\\
a){x^3} - 3{x^2} - 4x + 12\\
= {x^2}\left( {x - 3} \right) - 4\left( {x - 3} \right)\\
= \left( {x - 3} \right)\left( {{x^2} - 4} \right)\\
= \left( {x - 3} \right)\left( {x - 2} \right)\left( {x + 2} \right)\\
b)2{x^2} - 2{y^2} - 6x - 6y\\
= 2\left( {{x^2} - {y^2}} \right) - 6\left( {x + y} \right)\\
= 2\left( {x + y} \right)\left( {x - y} \right) - 6\left( {x + y} \right)\\
= 2\left( {x + y} \right)\left( {x - y - 3} \right)\\
c){x^3} + 3{x^2} - 3x - 1\\
= {x^3} - 1 + 3{x^2} - 3x\\
= \left( {x - 1} \right)\left( {{x^2} + x + 1} \right) + 3x\left( {x - 1} \right)\\
= \left( {x - 1} \right)\left( {{x^2} + x + 1 + 3x} \right)\\
= \left( {x - 1} \right)\left( {{x^2} + 4x + 1} \right)\\
d){x^4} - 5{x^2} + 4\\
= {x^4} - {x^2} - 4{x^2} + 4\\
= \left( {{x^2} - 1} \right)\left( {{x^2} - 4} \right)\\
= \left( {x - 1} \right)\left( {x + 1} \right)\left( {x - 2} \right)\left( {x + 2} \right)\\
B7\\
{x^2} - x + 1\\
= {x^2} - 2.x.\frac{1}{2} + \frac{1}{4} + \frac{3}{4}\\
= {\left( {x - \frac{1}{2}} \right)^2} + \frac{3}{4} \ge \frac{3}{4} > 0\\
Vậy\,{x^2} - x + 1 > 0\forall x \in R
\end{array}$
