Giải thích các bước giải:
$\begin{array}{l}
a)\\
a1)6x\left( {3x - 1} \right) - \left( {x + 3} \right)\left( {2x - 5} \right)\\
= 18{x^2} - 6x - \left( {2{x^2} + x - 15} \right)\\
= 16{x^2} - 7x + 15\\
a2)1 - \frac{2}{{x + 2}} + 3x + \frac{2}{{4 - {x^2}}} + \frac{2}{{x - 2}}\left( {DK:x \ne \pm 2} \right)\\
= 1 - \frac{2}{{x + 2}} + 3x - \frac{2}{{\left( {x - 2} \right)\left( {x + 2} \right)}} + \frac{2}{{x - 2}}\\
= 1 + 3x + 2\left( {\frac{1}{{x - 2}} - \frac{1}{{x + 2}} - \frac{1}{{\left( {x - 2} \right)\left( {x + 2} \right)}}} \right)\\
= 1 + 3x + 2.\frac{{x + 2 - \left( {x - 2} \right) - 1}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= 1 + 3x + \frac{6}{{{x^2} - 4}}\\
b)\\
b1){x^2}y - 5x + 2xy - 10\\
= \left( {{x^2}y + 2xy} \right) - 5\left( {x + 2} \right)\\
= \left( {x + 2} \right)\left( {xy - 5} \right)\\
b2)5{x^2} - {y^2} + {x^2} - 5xy\\
= \left( {5{x^2} - 5xy} \right) + \left( {{x^2} - {y^2}} \right)\\
= \left( {x - y} \right)\left( {5x + x + y} \right)\\
= \left( {x - y} \right)\left( {6x + y} \right)
\end{array}$
