Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
6,\\
1,\\
5{x^2} - 10xy + 5{y^2} - 20{z^2}\\
= 5.\left[ {\left( {{x^2} - 2xy + {y^2}} \right) - 4{z^2}} \right]\\
= 5.\left[ {{{\left( {x - y} \right)}^2} - {{\left( {2z} \right)}^2}} \right]\\
= 5.\left( {x - y - 2z} \right)\left( {x - y + 2z} \right)\\
2,\\
16x - 5{x^2} - 3\\
= - 5{x^2} + 16x - 3\\
= \left( { - 5{x^2} + 15x} \right) + \left( {x - 3} \right)\\
= - 5x.\left( {x - 3} \right) + \left( {x - 3} \right)\\
= \left( {x - 3} \right)\left( { - 5x + 1} \right)\\
3,\\
{x^2} - 5x + 5y - {y^2}\\
= \left( {{x^2} - {y^2}} \right) - \left( {5x - 5y} \right)\\
= \left( {x - y} \right)\left( {x + y} \right) - 5.\left( {x - y} \right)\\
= \left( {x - y} \right)\left( {x + y - 5} \right)\\
4,\\
3{x^2} - 6xy + 3{y^2} - 12{z^2}\\
= 3.\left[ {\left( {{x^2} - 2xy + {y^2}} \right) - 4{z^2}} \right]\\
= 3.\left[ {{{\left( {x - y} \right)}^2} - {{\left( {2z} \right)}^2}} \right]\\
= 3.\left( {x - y - 2z} \right).\left( {x - y + 2z} \right)\\
5,\\
{x^2} + 4x + 3 = \left( {{x^2} + x} \right) + \left( {3x + 3} \right)\\
= x.\left( {x + 1} \right) + 3.\left( {x + 1} \right)\\
= \left( {x + 1} \right)\left( {x + 3} \right)\\
6,\\
{\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}\\
7,\\
{x^2} - 4x - 5 = \left( {{x^2} - 5x} \right) + \left( {x - 5} \right)\\
= x\left( {x - 5} \right) + \left( {x - 5} \right) = \left( {x - 5} \right)\left( {x + 1} \right)\\
8,\\
{x^5} - 3{x^4} + 3{x^3} - {x^2} = {x^2}.\left( {{x^3} - 3{x^2} + 3x - 1} \right) = {x^2}{\left( {x - 1} \right)^3}\\
10,\\
1.\\
A = 4x - {x^2} + 3\\
= \left( { - {x^2} + 4x - 4} \right) + 7\\
= 7 - \left( {{x^2} - 4x + 4} \right)\\
= 7 - {\left( {x - 2} \right)^2} \le 7,\,\,\,\forall x\\
\Rightarrow {A_{\max }} = 7 \Leftrightarrow {\left( {x - 2} \right)^2} = 0 \Leftrightarrow x = 2\\
2.\\
B = - {x^2} + 6x - 11\\
= \left( { - {x^2} + 6x - 9} \right) - 2\\
= - \left( {{x^2} - 6x + 9} \right) - 2\\
= - 2 - {\left( {x - 3} \right)^2} \le - 2,\,\,\,\forall x\\
\Rightarrow {B_{\max }} = - 2 \Leftrightarrow {\left( {x - 3} \right)^2} = 0 \Leftrightarrow x = 3
\end{array}\)
