Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
12,\\
P = {x^2} - 4xy + 4{y^2} - 4\\
= \left( {{x^2} - 4xy + 4{y^2}} \right) - 4\\
= {\left( {x - 2y} \right)^2} - {2^2}\\
= \left( {x - 2y - 2} \right)\left( {x - 2y + 2} \right)\\
13,\\
Q = 16 - {x^2} + 2xy - {y^2}\\
= 16 - \left( {{x^2} - 2xy + {y^2}} \right)\\
= {4^2} - {\left( {x - y} \right)^2}\\
= \left[ {4 - \left( {x - y} \right)} \right].\left[ {4 + \left( {x - y} \right)} \right]\\
= \left( {4 - x + y} \right)\left( {4 + x - y} \right)\\
14,\\
M = {x^2} + 2x - 4{y^2} - 4y\\
= \left( {{x^2} - 4{y^2}} \right) + \left( {2x - 4y} \right)\\
= \left[ {{x^2} - {{\left( {2y} \right)}^2}} \right] + 2.\left( {x - 2y} \right)\\
= \left( {x - 2y} \right)\left( {x + 2y} \right) + 2.\left( {x - 2y} \right)\\
= \left( {x - 2y} \right)\left( {x + 2y + 2} \right)\\
15,\\
N = 6{x^2} - 5x + 1\\
= \left( {6{x^2} - 2x} \right) - \left( {3x - 1} \right)\\
= 2x.\left( {3x - 1} \right) - \left( {3x - 1} \right)\\
= \left( {3x - 1} \right)\left( {2x - 1} \right)\\
16,\\
O = 3{x^2} + 13x - 10\\
= \left( {3{x^2} + 15x} \right) - \left( {2x + 10} \right)\\
= 3x.\left( {x + 5} \right) - 2.\left( {x + 5} \right)\\
= \left( {x + 5} \right)\left( {3x - 2} \right)\\
17,\\
Z = {\left( {{x^2} - 4x} \right)^2} + 8\left( {{x^2} - 4x} \right) + 15\\
= \left[ {{{\left( {{x^2} - 4x} \right)}^2} + 3.\left( {{x^2} - 4x} \right)} \right] + \left[ {5.\left( {{x^2} - 4x} \right) + 15} \right]\\
= \left( {{x^2} - 4x} \right).\left[ {\left( {{x^2} - 4x} \right) + 3} \right] + 5.\left[ {\left( {{x^2} - 4x} \right) + 3} \right]\\
= \left[ {\left( {{x^2} - 4x} \right) + 3} \right].\left[ {\left( {{x^2} - 4x} \right) + 5} \right]\\
= \left( {{x^2} - 4x + 3} \right).\left( {{x^2} - 4x + 5} \right)\\
= \left[ {\left( {{x^2} - x} \right) - \left( {3x - 3} \right)} \right].\left( {{x^2} - 4x + 5} \right)\\
= \left[ {x.\left( {x - 1} \right) - 3.\left( {x - 1} \right)} \right].\left( {{x^2} - 4x + 5} \right)\\
= \left( {x - 1} \right)\left( {x - 3} \right)\left( {{x^2} - 4x + 5} \right)
\end{array}\)
