Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
2,\\
M = {x^2} - 4xy + 5{y^2} - 2y + 3\\
= \left( {{x^2} - 4xy + 4{y^2}} \right) + \left( {{y^2} - 2y + 1} \right) + 2\\
= \left( {{x^2} - 2.x.2y + {{\left( {2y} \right)}^2}} \right) + {\left( {y - 1} \right)^2} + 2\\
= {\left( {x - 2y} \right)^2} + {\left( {y - 1} \right)^2} + 2 \ge 2 > 0,\,\,\,\forall x,y\\
3,\\
a,\\
{a^3} - {a^2}c + {a^2}b - abc\\
= \left( {{a^3} - {a^2}c} \right) + \left( {{a^2}b - abc} \right)\\
= {a^2}\left( {a - c} \right) + ab\left( {a - c} \right)\\
= \left( {a - c} \right).\left( {{a^2} + ab} \right)\\
= a.\left( {a - c} \right)\left( {a + b} \right)\\
b,\\
{\left( {{x^2} + 1} \right)^2} - 4{x^2}\\
= {\left( {{x^2} + 1} \right)^2} - {\left( {2x} \right)^2}\\
= \left( {{x^2} + 1 - 2x} \right).\left( {{x^2} + 1 + 2x} \right)\\
= {\left( {x - 1} \right)^2}.{\left( {x + 1} \right)^2}\\
c,\\
{x^2} - 10x - 9{y^2} + 25\\
= \left( {{x^2} - 10x + 25} \right) - 9{y^2}\\
= {\left( {x - 5} \right)^2} - {\left( {3y} \right)^2}\\
= \left( {x - 5 - 3y} \right)\left( {x - 5 + 3y} \right)\\
d,\\
4{x^2} - 36x + 56\\
= 4.\left( {{x^2} - 9x + 14} \right)\\
= 4.\left[ {\left( {{x^2} - 2x} \right) - \left( {7x - 14} \right)} \right]\\
= 4.\left[ {x.\left( {x - 2} \right) - 7.\left( {x - 2} \right)} \right]\\
= 4.\left( {x - 2} \right)\left( {x - 7} \right)
\end{array}\)
