Đáp án:

\(\begin{array}{l}
1,\\
a,\\
\left( {x - 2y - 4} \right)\left( {x - 2y + 4} \right)\\
b,\\
\left( {x - 1} \right).{x^2}\left( {{x^2} + 1} \right)\\
c,\\
\left( {x - 2} \right)\left( {x - 4} \right)\\
d,\\
{\left( {x - 1} \right)^2}\left( {x + 1} \right)\left( {{x^2} + x + 1} \right)\\
e,\\
\left( {x - 1} \right)\left( {x - y} \right)\left( {x + y} \right)\\
f,\\
\left( {x + y} \right)\left( {3 - x - y} \right)\\
g,\\
\left( {x + 1} \right)\left( {x + 3} \right)\\
h,\\
\left( {2x - 1} \right)\left( {2x + 3} \right)\\
m,\\
\left( {x - 4} \right)\left( {x + 3} \right)\\
2,\\
a,\\
\left[ \begin{array}{l}
x = 0\\
x = 16
\end{array} \right.\\
b,\\
\left[ \begin{array}{l}
x = 2\\
x = 0
\end{array} \right.\\
c,\\
\left[ \begin{array}{l}
x = 8\\
x =  - \dfrac{2}{3}
\end{array} \right.\\
d,\\
\left[ \begin{array}{l}
x = 1\\
x = 2
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
a,\\
{x^2} - 16 - 4xy + 4{y^2}\\
 = \left( {{x^2} - 4xy + 4{y^2}} \right) - 16\\
 = \left[ {{x^2} - 2.x.2y + {{\left( {2y} \right)}^2}} \right] - {4^2}\\
 = {\left( {x - 2y} \right)^2} - {4^2}\\
 = \left( {x - 2y - 4} \right)\left( {x - 2y + 4} \right)\\
b,\\
{x^5} - {x^4} + {x^3} - {x^2}\\
 = \left( {{x^5} - {x^4}} \right) + \left( {{x^3} - {x^2}} \right)\\
 = {x^4}.\left( {x - 1} \right) + {x^2}\left( {x - 1} \right)\\
 = \left( {x - 1} \right)\left( {{x^4} + {x^2}} \right)\\
 = \left( {x - 1} \right).{x^2}\left( {{x^2} + 1} \right)\\
c,\\
{x^2} - 6x + 8\\
 = \left( {{x^2} - 2x} \right) + \left( { - 4x + 8} \right)\\
 = x\left( {x - 2} \right) - 4.\left( {x - 2} \right)\\
 = \left( {x - 2} \right)\left( {x - 4} \right)\\
d,\\
{x^5} - {x^3} - {x^2} + 1\\
 = \left( {{x^5} - {x^3}} \right) - \left( {{x^2} - 1} \right)\\
 = {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}\\
 = \left( {{x^3} - {x^2}} \right) - \left( {x{y^2} - {y^2}} \right)\\
 = {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}\\
 = \left( {3x + 3y} \right) - \left( {{x^2} + 2xy + {y^2}} \right)\\
 = 3.\left( {x + y} \right) - {\left( {x + y} \right)^2}\\
 = \left( {x + y} \right).\left[ {3 - \left( {x + y} \right)} \right]\\
 = \left( {x + y} \right)\left( {3 - x - y} \right)\\
g,\\
{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)\\
h,\\
4{x^2} + 4x - 3\\
 = \left( {4{x^2} - 2x} \right) + \left( {6x - 3} \right)\\
 = 2x.\left( {2x - 1} \right) + 3.\left( {2x - 1} \right)\\
 = \left( {2x - 1} \right)\left( {2x + 3} \right)\\
m,\\
{x^2} - x - 12\\
 = \left( {{x^2} - 4x} \right) + \left( {3x - 12} \right)\\
 = x\left( {x - 4} \right) + 3.\left( {x - 4} \right)\\
 = \left( {x - 4} \right)\left( {x + 3} \right)\\
2,\\
a,\\
{x^2} - 16x = 0\\
 \Leftrightarrow x\left( {x - 16} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x - 16 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 16
\end{array} \right.\\
b,\\
{x^4} - 2{x^3} + 10{x^2} - 20x = 0\\
 \Leftrightarrow \left( {{x^4} - 2{x^3}} \right) + \left( {10{x^2} - 20x} \right) = 0\\
 \Leftrightarrow {x^3}.\left( {x - 2} \right) + 10x\left( {x - 2} \right) = 0\\
 \Leftrightarrow \left( {x - 2} \right)\left( {{x^3} + 10x} \right) = 0\\
 \Leftrightarrow \left( {x - 2} \right).x.\left( {{x^2} + 10} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
x = 0\\
{x^2} + 10 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = 0\\
{x^2} =  - 10\,\,\,\left( {L,\,\,{x^2} \ge 0} \right)
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = 2\\
x = 0
\end{array} \right.\\
c,\\
{\left( {2x - 3} \right)^2} = {\left( {x + 5} \right)^2}\\
 \Leftrightarrow {\left( {2x - 3} \right)^2} - {\left( {x + 5} \right)^2} = 0\\
 \Leftrightarrow \left[ {\left( {2x - 3} \right) - \left( {x + 5} \right)} \right].\left[ {\left( {2x - 3} \right) + \left( {x + 5} \right)} \right] = 0\\
 \Leftrightarrow \left( {2x - 3 - x - 5} \right).\left( {2x - 3 + x + 5} \right) = 0\\
 \Leftrightarrow \left( {x - 8} \right).\left( {3x + 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 8 = 0\\
3x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 8\\
x =  - \dfrac{2}{3}
\end{array} \right.\\
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 {x^2}\left( {x - 1} \right) - 4.{\left( {x - 1} \right)^2} = 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 \left( {x - 1} \right).\left( {{x^2} - 2.x.2 + {2^2}} \right) = 0\\
 \Leftrightarrow \left( {x - 1} \right){\left( {x - 2} \right)^2} = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
{\left( {x - 2} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = 2
\end{array} \right.
\end{array}\)