Giải thích các bước giải:

$\begin{array}{l}
B1:\\
a)4x\left( {x - 3} \right) - 3x\left( {2 + x} \right)\\
 = 4{x^2} - 12x - 6x - 3{x^2}\\
 = {x^2} - 18x\\
b)2x\left( {5x + 2} \right) + \left( {2x - 3} \right)\left( {3x - 1} \right)\\
 = 10{x^2} + 4x + 6{x^2} - 11x + 3\\
 = 16{x^2} - 7x + 3\\
c){\left( {x - 1} \right)^2} - \left( {x + 2} \right)\left( {x - 2} \right)\\
 = {x^2} - 2x + 1 - {x^2} + 4\\
 =  - 2x + 5\\
B2:\\
a)2x\left( {x - y} \right) + 3\left( {y - x} \right)\\
 = \left( {x - y} \right)\left( {2x - 3} \right)\\
b){a^2} - 25 - 2ab + {b^2}\\
 = \left( {{a^2} - 2ab + {b^2}} \right) - {5^2}\\
 = {\left( {a - b} \right)^2} - {5^2}\\
 = \left( {a - b - 5} \right)\left( {a - b + 5} \right)\\
c)5x - 5y + {x^2} - 2xy + {y^2}\\
 = 5\left( {x - y} \right) + {\left( {x - y} \right)^2}\\
 = \left( {x - y} \right)\left( {x - y + 5} \right)\\
d){x^2} - 5x + 6\\
 = {x^2} - 2x - 3x + 6\\
 = \left( {x - 2} \right)\left( {x - 3} \right)\\
e){x^2} - 6xy + 9{y^2} - 25{z^2}\\
 = {\left( {x - 3y} \right)^2} - {\left( {5z} \right)^2}\\
 = \left( {x - 3y - 5z} \right)\left( {x - 3y + 5z} \right)\\
f)3{x^2} - 3{y^2} - 12x + 12y\\
 = 3\left( {x - y} \right)\left( {x + y} \right) - 12\left( {x - y} \right)\\
 = 3\left( {x - y} \right)\left( {x + y - 4} \right)\\
g)4{x^3} + 4x{y^2} + 8{x^2}y - 16x\\
 = 4x\left( {{x^2} + 2xy + {y^2} - 4} \right)\\
 = 4x\left( {{{\left( {x + y} \right)}^2} - 4} \right)\\
 = 4x\left( {x + y - 2} \right)\left( {x + y + 2} \right)\\
h){x^2} - 5x + 4\\
 = {x^2} - x - 4x + 4\\
 = \left( {x - 1} \right)\left( {x - 4} \right)\\
i){x^4} - 5{x^2} + 4\\
 = {x^4} - {x^2} - 4{x^2} + 4\\
 = \left( {{x^2} - 1} \right)\left( {{x^2} - 4} \right)\\
 = \left( {x - 1} \right)\left( {x + 1} \right)\left( {x - 2} \right)\left( {x + 2} \right)
\end{array}$