Đáp án:
$\begin{array}{l}
1a){x^2} - 16 - 4xy + 4{y^2}\\
= {x^2} - 4xy + 4{y^2} - 16\\
= {\left( {x - 2y} \right)^2} - 16\\
= \left( {x - 2y - 4} \right)\left( {x - 2y + 4} \right)\\
b){x^5} - {x^4} + {x^3} - {x^2}\\
= {x^4}\left( {x - 1} \right) + {x^2}\left( {x - 1} \right)\\
= {x^2}\left( {x - 1} \right)\left( {{x^2} + 1} \right)\\
c){x^2} - 6x + 8\\
= {x^2} - 2x - 4x + 8\\
= \left( {x - 2} \right)\left( {x - 4} \right)\\
d){x^5} - {x^3} - {x^2} + 1\\
= {x^3}\left( {{x^2} - 1} \right) - \left( {{x^2} - 1} \right)\\
= \left( {{x^2} - 1} \right)\left( {{x^3} - 1} \right)\\
= \left( {x + 1} \right)\left( {x - 1} \right)\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)\\
= {\left( {x - 1} \right)^2}\left( {x + 1} \right)\left( {{x^2} + x + 1} \right)\\
e){x^3} - {x^2} - x{y^2} + {y^2}\\
= {x^2}\left( {x - 1} \right) - {y^2}\left( {x - 1} \right)\\
= \left( {x - 1} \right)\left( {{x^2} - {y^2}} \right)\\
= \left( {x - 1} \right)\left( {x - y} \right)\left( {x + y} \right)\\
f)3x + 3y - {x^2} - 2xy - {y^2}\\
= 3\left( {x + y} \right) - \left( {{x^2} + 2xy + {y^2}} \right)\\
= 3\left( {x + y} \right) - {\left( {x + y} \right)^2}\\
= \left( {x + y} \right)\left( {3 - x - y} \right)\\
g){x^2} + 4x + 3\\
= \left( {x + 1} \right)\left( {x + 3} \right)\\
h)4{x^2} + 4x - 3\\
= 4{x^2} + 4x + 1 - 4\\
= {\left( {2x + 1} \right)^2} - 4\\
= \left( {2x + 1 - 2} \right)\left( {2x + 1 + 2} \right)\\
= \left( {2x - 1} \right)\left( {2x + 3} \right)\\
m){x^2} - x - 12\\
= \left( {x - 4} \right)\left( {x + 3} \right)\\
n)4{x^4} + 4{x^2}{y^2} + 8{y^4}\\
= 4\left( {{x^4} + {x^2}{y^2} + 2{y^4}} \right)\\
2)a){x^2} - 16x = 0\\
\Leftrightarrow x\left( {x - 16} \right) = 0\\
\Leftrightarrow x = 0/x = 16\\
Vậy\,x = 0;x = 16\\
b){x^4} - 2{x^3} + 10{x^2} - 20x = 0\\
\Leftrightarrow {x^3}\left( {x - 2} \right) + 10x\left( {x - 2} \right) = 0\\
\Leftrightarrow x\left( {x - 2} \right)\left( {{x^2} + 10} \right) = 0\\
\Leftrightarrow x = 0/x = 2\\
Vậy\,x = 0;x = 2\\
c){\left( {2x - 3} \right)^2} = {\left( {x + 5} \right)^2}\\
\Leftrightarrow \left[ \begin{array}{l}
2x - 3 = x + 5\\
2x - 3 = - x - 5
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 8\\
x = \dfrac{{ - 2}}{3}
\end{array} \right.\\
Vậy\,x = 8;x = - \dfrac{2}{3}\\
d){x^2}\left( {x - 1} \right) - 4{x^2} + 8x - 4 = 0\\
\Leftrightarrow {x^2}\left( {x - 1} \right) - 4\left( {{x^2} - 2x + 1} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {{x^2} - 4\left( {x - 1} \right)} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {{x^2} - 4x + 4} \right) = 0\\
\Leftrightarrow x = 1;x = 2\\
Vậy\,x = 1;x = 2
\end{array}$
