Tính Bằng hằng đẳng thức (x-3) ( x^2 + 3x +9) (x -5) ( x^2 +5x +25) x^2 + 4x +4 x^2 – 8x +16 ( x-5) (x+ 5) x^3 + 12x^2 +48x +64 x^3 – 6x^2 + 12x -8

Đáp án:

\(11){x^2} – 1 = \left( {x – 1} \right)\left( {x + 1} \right)\)

Giải thích các bước giải:

\(\begin{array}{l}
1)(x – 3)({x^2} + 3x + 9)\\
 = {x^3} – {3^3} = {x^3} – 27\\
2)(x – 5)({x^2} + 5x + 25)\\
 = {x^3} – {5^3} = {x^3} – 125\\
3){x^2} + 4x + 4 = {\left( {x + 2} \right)^2}\\
4){x^2} – 8x + 16 = {\left( {x – 4} \right)^2}\\
5)\left( {{\rm{ }}x – 5} \right)\left( {x + {\rm{ }}5} \right) = {x^2} – {5^2} = {x^2} – 25\\
6){x^3} + 12{x^2} + 48x + 64\\
 = {x^3} + 3.{x^2}.4 + 3.16.x + {4^3}\\
 = {\left( {x + 3} \right)^3}\\
7){x^3} – 6{x^2} + 12x – 8\\
 = {x^3} – 3.{x^2}.2 + 3.4.x – {2^3}\\
 = {\left( {x – 2} \right)^3}\\
8)(x + 2)({x^2} – 2x + 4)\\
 = {x^3} + {2^3} = {x^3} + 8\\
9)(x – 3)({x^2} + 3x + 9)\\
 = {x^2} – 27\\
10){x^2} + 2x + 1\\
 = {\left( {x + 1} \right)^2}\\
11){x^2} – 1 = \left( {x – 1} \right)\left( {x + 1} \right)
\end{array}\)