Đáp án:
\(11){x^2} - 1 = \left( {x - 1} \right)\left( {x + 1} \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
1)(x - 3)({x^2} + 3x + 9)\\
= {x^3} - {3^3} = {x^3} - 27\\
2)(x - 5)({x^2} + 5x + 25)\\
= {x^3} - {5^3} = {x^3} - 125\\
3){x^2} + 4x + 4 = {\left( {x + 2} \right)^2}\\
4){x^2} - 8x + 16 = {\left( {x - 4} \right)^2}\\
5)\left( {{\rm{ }}x - 5} \right)\left( {x + {\rm{ }}5} \right) = {x^2} - {5^2} = {x^2} - 25\\
6){x^3} + 12{x^2} + 48x + 64\\
= {x^3} + 3.{x^2}.4 + 3.16.x + {4^3}\\
= {\left( {x + 3} \right)^3}\\
7){x^3} - 6{x^2} + 12x - 8\\
= {x^3} - 3.{x^2}.2 + 3.4.x - {2^3}\\
= {\left( {x - 2} \right)^3}\\
8)(x + 2)({x^2} - 2x + 4)\\
= {x^3} + {2^3} = {x^3} + 8\\
9)(x - 3)({x^2} + 3x + 9)\\
= {x^2} - 27\\
10){x^2} + 2x + 1\\
= {\left( {x + 1} \right)^2}\\
11){x^2} - 1 = \left( {x - 1} \right)\left( {x + 1} \right)
\end{array}\)
