Đáp án:
$\begin{array}{l}
1)\\
{x^2} - 5x + 6\\
= {x^2} - 2.x.\dfrac{5}{2} + \dfrac{{25}}{4} - \dfrac{1}{4}\\
= {\left( {x - \dfrac{5}{2}} \right)^2} - \dfrac{1}{4}\\
= \left( {x - \dfrac{5}{2} - \dfrac{1}{2}} \right)\left( {x - \dfrac{5}{2} + \dfrac{1}{2}} \right)\\
= \left( {x - 3} \right)\left( {x - 2} \right)\\
2)\\
{x^2} + 5x + 6\\
= {\left( {x + \dfrac{5}{2}} \right)^2} - \dfrac{1}{4}\\
= \left( {x + \dfrac{5}{2} + \dfrac{1}{2}} \right)\left( {x + \dfrac{5}{2} - \dfrac{1}{2}} \right)\\
= \left( {x + 3} \right)\left( {x + 2} \right)\\
3){x^2} - 7x + 12\\
= {x^2} - 2.x.\dfrac{7}{2} + \dfrac{{49}}{4} - \dfrac{1}{4}\\
= {\left( {x - \dfrac{7}{2}} \right)^2} - \dfrac{1}{4}\\
= \left( {x - \dfrac{7}{2} + \dfrac{1}{2}} \right)\left( {x - \dfrac{7}{2} - \dfrac{1}{2}} \right)\\
= \left( {x - 3} \right)\left( {x - 4} \right)\\
4){x^2} + 7x + 12\\
= \left( {x + 3} \right)\left( {x + 4} \right)\\
5){x^2} + x - 12\\
= {x^2} + 2.x.\dfrac{1}{2} + \dfrac{1}{4} - \dfrac{{49}}{4}\\
= {\left( {x + \dfrac{1}{2}} \right)^2} - \dfrac{{49}}{4}\\
= \left( {x + \dfrac{1}{2} + \dfrac{7}{2}} \right)\left( {x + \dfrac{1}{2} - \dfrac{7}{2}} \right)\\
= \left( {x + 4} \right)\left( {x - 3} \right)\\
5){x^2} - x - 12\\
= \left( {x - 4} \right)\left( {x + 3} \right)\\
6){x^2} - 9x + 20\\
= {x^2} - 2.x.\dfrac{9}{2} + \dfrac{{81}}{4} - \dfrac{1}{4}\\
= {\left( {x - \dfrac{9}{2}} \right)^2} - \dfrac{1}{4}\\
= \left( {x - \dfrac{9}{2} - \dfrac{1}{2}} \right)\left( {x - \dfrac{9}{2} + \dfrac{1}{2}} \right)\\
= \left( {x - 5} \right)\left( {x - 4} \right)\\
7){x^2} + 9x + 20\\
= \left( {x + 5} \right)\left( {x + 4} \right)\\
8){x^2} + x - 20\\
= {x^2} + 2.x.\dfrac{1}{2} + \dfrac{1}{4} - \dfrac{{81}}{4}\\
= {\left( {x + \dfrac{1}{2}} \right)^2} - \dfrac{{81}}{4}\\
= \left( {x + \dfrac{1}{2} + \dfrac{9}{2}} \right)\left( {x + \dfrac{1}{2} - \dfrac{9}{2}} \right)\\
= \left( {x + 5} \right)\left( {x - 4} \right)\\
10){x^2} - x - 20\\
= \left( {x - 5} \right)\left( {x + 4} \right)
\end{array}$
