Đáp án:

\(\begin{array}{l}
1,\\
a,\\
\left( {x - y} \right)\left( {x + 1} \right)\\
b,\\
\left( {x + y} \right)\left( {z - 5} \right)\\
c,\\
\left( {x - y} \right)\left( {3x - 5} \right)\\
d,\\
\left( {x - 3} \right).\left( {x - 2} \right)\left( {x + 2} \right)\\
e,\\
\left( {x - 5} \right)\left( {x - 3} \right)\left( {x + 3} \right)\\
f,\\
{\left( {x + 1} \right)^2}.\left( {{x^2} - x + 1} \right)\\
2,\\
1,\\
\left( {x + y} \right).\left( {{x^2} - xy + {y^2} - 1} \right)\\
2,\\
\left( {x - y - 2z} \right).\left( {x - y + 2z} \right)\\
3,\\
\left( {x - 1} \right).\left( {x + y} \right)\\
4,\\
{\left( {x + 2y} \right)^3}\\
5,\\
\left( {x - y} \right).\left( {x - y - z} \right)\\
6,\\
\left( {x - y} \right)\left( {x + y - 1} \right)\\
7,\\
3.\left( {x + y - z} \right)\left( {x + y + z} \right)\\
8,\\
\left( {x - y - z + t} \right).\left( {x - y + z - t} \right)\\
9,\\
\left( {{x^2} - xy + {y^2}} \right).\left( {x + y + 1} \right)\\
10,\\
\left( {x + 3} \right)\left( {x - 9} \right)\\
3,\\
a,\\
A = \left( {2x + 1 - y} \right)\left( {2x + 1 + y} \right)\\
A = 416\\
b,\\
B = \left( {x - y - 1} \right)\left( {x + y + 1} \right)\\
B = 8600\\
4,\\
a,\\
\left[ \begin{array}{l}
x + 3 = 0\\
2 - x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x =  - 3\\
x = 2
\end{array} \right.\\
b,\\
x = \dfrac{5}{2}\\
c,\\
\left[ \begin{array}{l}
x =  - 3\\
x = 0\\
x = 2
\end{array} \right.\\
d,\,\\
\left[ \begin{array}{l}
x = 3\\
x = 2\\
x =  - 2
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
1,\\
a,\\
{x^2} - xy + x - y\\
 = \left( {{x^2} - xy} \right) + \left( {x - y} \right)\\
 = x\left( {x - y} \right) + \left( {x - y} \right)\\
 = \left( {x - y} \right)\left( {x + 1} \right)\\
b,\\
xz + yz - 5x - 5y\\
 = \left( {xz + yz} \right) - \left( {5x + 5y} \right)\\
 = z\left( {x + y} \right) - 5\left( {x + y} \right)\\
 = \left( {x + y} \right)\left( {z - 5} \right)\\
c,\\
3{x^2} - 3xy - 5x + 5y\\
 = \left( {3{x^2} - 3xy} \right) + \left( { - 5x + 5y} \right)\\
 = 3x.\left( {x - y} \right) - 5.\left( {x - y} \right)\\
 = \left( {x - y} \right)\left( {3x - 5} \right)\\
d,\\
{x^3} - 3{x^2} - 4x + 12\\
 = \left( {{x^3} - 3{x^2}} \right) + \left( { - 4x + 12} \right)\\
 = {x^2}.\left( {x - 3} \right) - 4.\left( {x - 3} \right)\\
 = \left( {x - 3} \right).\left( {{x^2} - 4} \right)\\
 = \left( {x - 3} \right).\left( {{x^2} - {2^2}} \right)\\
 = \left( {x - 3} \right).\left( {x - 2} \right)\left( {x + 2} \right)\\
e,\\
45 + {x^3} - 5{x^2} - 9x\\
 = \left( {{x^3} - 5{x^2}} \right) + \left( { - 9x + 45} \right)\\
 = {x^2}\left( {x - 5} \right) - 9.\left( {x - 5} \right)\\
 = \left( {x - 5} \right).\left( {{x^2} - 9} \right)\\
 = \left( {x - 5} \right)\left( {{x^2} - {3^2}} \right)\\
 = \left( {x - 5} \right)\left( {x - 3} \right)\left( {x + 3} \right)\\
f,\\
{x^4} + {x^3} + x + 1\\
 = \left( {{x^4} + {x^3}} \right) + \left( {x + 1} \right)\\
 = {x^3}\left( {x + 1} \right) + \left( {x + 1} \right)\\
 = \left( {x + 1} \right)\left( {{x^3} + 1} \right)\\
 = \left( {x + 1} \right).\left( {{x^3} + {1^3}} \right)\\
 = \left( {x + 1} \right).\left( {x + 1} \right).\left( {{x^2} - x.1 + {1^2}} \right)\\
 = {\left( {x + 1} \right)^2}.\left( {{x^2} - x + 1} \right)\\
2,\\
1,\\
{x^3} - x + {y^3} - y\\
 = \left( {{x^3} + {y^3}} \right) + \left( { - x - y} \right)\\
 = \left( {x + y} \right).\left( {{x^2} - xy + {y^2}} \right) - \left( {x + y} \right)\\
 = \left( {x + y} \right).\left( {{x^2} - xy + {y^2} - 1} \right)\\
2,\\
{x^2} - 2xy - 4{z^2} + {y^2}\\
 = \left( {{x^2} - 2xy + {y^2}} \right) - 4{z^2}\\
 = {\left( {x - y} \right)^2} - {\left( {2z} \right)^2}\\
 = \left[ {\left( {x - y} \right) - 2z} \right].\left[ {\left( {x - y} \right) + 2z} \right]\\
 = \left( {x - y - 2z} \right).\left( {x - y + 2z} \right)\\
3,\\
x\left( {x - 1} \right) - y.\left( {1 - x} \right)\\
 = x\left( {x - 1} \right) + y.\left( {x - 1} \right)\\
 = \left( {x - 1} \right).\left( {x + y} \right)\\
4,\\
{x^3} + 6{x^2}y + 12x{y^2} + 8{y^3}\\
 = {x^3} + 3.{x^2}.2y + 3.x.4{y^2} + 8{y^3}\\
 = {x^3} + 3.{x^2}.2y + 3.x.{\left( {2y} \right)^2} + {\left( {2y} \right)^3}\\
 = {\left( {x + 2y} \right)^3}\\
5,\\
{x^2} - 2xy + {y^2} - xz + yz\\
 = \left( {{x^2} - 2xy + {y^2}} \right) + \left( { - xz + yz} \right)\\
 = {\left( {x - y} \right)^2} - z.\left( {x - y} \right)\\
 = \left( {x - y} \right).\left[ {\left( {x - y} \right) - z} \right]\\
 = \left( {x - y} \right).\left( {x - y - z} \right)\\
6,\\
{x^2} - {y^2} - x + y\\
 = \left( {{x^2} - {y^2}} \right) + \left( { - x + y} \right)\\
 = \left( {x - y} \right)\left( {x + y} \right) - \left( {x - y} \right)\\
 = \left( {x - y} \right).\left[ {\left( {x + y} \right) - 1} \right]\\
 = \left( {x - y} \right)\left( {x + y - 1} \right)\\
7,\\
3{x^2} + 6xy + 3{y^2} - 3{z^2}\\
 = 3.\left( {{x^2} + 2xy + {y^2} - {z^2}} \right)\\
 = 3.\left[ {\left( {{x^2} + 2xy + {y^2}} \right) - {z^2}} \right]\\
 = 3.\left[ {{{\left( {x + y} \right)}^2} - {z^2}} \right]\\
 = 3.\left[ {\left( {x + y} \right) - z} \right].\left[ {\left( {x + y} \right) + z} \right]\\
 = 3.\left( {x + y - z} \right)\left( {x + y + z} \right)\\
8,\\
{x^2} - 2xy + {y^2} - {z^2} + 2zt - {t^2}\\
 = \left( {{x^2} - 2xy + {y^2}} \right) - \left( {{z^2} - 2zt + {t^2}} \right)\\
 = {\left( {x - y} \right)^2} - {\left( {z - t} \right)^2}\\
 = \left[ {\left( {x - y} \right) - \left( {z - t} \right)} \right].\left[ {\left( {x - y} \right) + \left( {z - t} \right)} \right]\\
 = \left( {x - y - z + t} \right).\left( {x - y + z - t} \right)\\
9,\\
{x^3} + {x^2} - xy + {y^2} + {y^3}\\
 = \left( {{x^3} + {y^3}} \right) + \left( {{x^2} - xy + {y^2}} \right)\\
 = \left( {x + y} \right).\left( {{x^2} - xy + {y^2}} \right) + \left( {{x^2} - xy + {y^2}} \right)\\
 = \left( {{x^2} - xy + {y^2}} \right).\left[ {\left( {x + y} \right) + 1} \right]\\
 = \left( {{x^2} - xy + {y^2}} \right).\left( {x + y + 1} \right)\\
10,\\
{x^2} - 6.\left( {x + 3} \right) - 9\\
 = \left( {{x^2} - 9} \right) - 6\left( {x + 3} \right)\\
 = \left( {{x^2} - {3^2}} \right) - 6.\left( {x + 3} \right)\\
 = \left( {x - 3} \right)\left( {x + 3} \right) - 6.\left( {x + 3} \right)\\
 = \left( {x + 3} \right).\left[ {\left( {x - 3} \right) - 6} \right]\\
 = \left( {x + 3} \right)\left( {x - 9} \right)\\
3,\\
a,\\
A = 4{x^2} - {y^2} + 4x + 1\\
 = \left( {4{x^2} + 4x + 1} \right) - {y^2}\\
 = \left[ {{{\left( {2x} \right)}^2} + 2.2x.1 + {1^2}} \right] - {y^2}\\
 = {\left( {2x + 1} \right)^2} - {y^2}\\
 = \left[ {\left( {2x + 1} \right) - y} \right].\left[ {\left( {2x + 1} \right) + y} \right]\\
 = \left( {2x + 1 - y} \right)\left( {2x + 1 + y} \right)\\
x = 10;\,y = 5 \Rightarrow A = \left( {2.10 + 1 - 5} \right).\left( {2.10 + 1 + 5} \right) = 16.26 = 416\\
b,\\
B = {x^2} - {y^2} - 2y - 1\\
 = {x^2} - \left( {{y^2} + 2y + 1} \right)\\
 = {x^2} - {\left( {y + 1} \right)^2}\\
 = \left[ {x - \left( {y + 1} \right)} \right].\left[ {x + \left( {y + 1} \right)} \right]\\
 = \left( {x - y - 1} \right)\left( {x + y + 1} \right)\\
x = 93;\,\,y = 6 \Rightarrow B = \left( {93 - 6 - 1} \right).\left( {93 + 6 + 1} \right) = 86.100 = 8600\\
4,\\
a,\\
2\left( {x + 3} \right) - {x^2} - 3x = 0\\
 \Leftrightarrow 2.\left( {x + 3} \right) - \left( {{x^2} + 3x} \right) = 0\\
 \Leftrightarrow 2.\left( {x + 3} \right) - x.\left( {x + 3} \right) = 0\\
 \Leftrightarrow \left( {x + 3} \right)\left( {2 - x} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 3 = 0\\
2 - x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x =  - 3\\
x = 2
\end{array} \right.\\
b,\\
4{x^2} - 25 - \left( {2x - 5} \right)\left( {2x + 7} \right) = 0\\
 \Leftrightarrow {\left( {2x} \right)^2} - {5^2} - \left( {2x - 5} \right)\left( {2x + 7} \right) = 0\\
 \Leftrightarrow \left( {2x - 5} \right)\left( {2x + 5} \right) - \left( {2x - 5} \right)\left( {2x + 7} \right) = 0\\
 \Leftrightarrow \left( {2x - 5} \right).\left[ {\left( {2x + 5} \right) - \left( {2x + 7} \right)} \right] = 0\\
 \Leftrightarrow \left( {2x - 5} \right).\left( { - 2} \right) = 0\\
 \Leftrightarrow 2x - 5 = 0\\
 \Leftrightarrow x = \dfrac{5}{2}\\
c,\\
{x^3} + 27 + \left( {x + 3} \right)\left( {x - 9} \right) = 0\\
 \Leftrightarrow {x^3} + {3^3} + \left( {x + 3} \right)\left( {x - 9} \right) = 0\\
 \Leftrightarrow \left( {x + 3} \right)\left( {{x^2} - x.3 + {3^2}} \right) + \left( {x + 3} \right)\left( {x - 9} \right) = 0\\
 \Leftrightarrow \left( {x + 3} \right)\left( {{x^2} - 3x + 9} \right) + \left( {x + 3} \right)\left( {x - 9} \right) = 0\\
 \Leftrightarrow \left( {x + 3} \right).\left[ {\left( {{x^2} - 3x + 9} \right) + \left( {x - 9} \right)} \right] = 0\\
 \Leftrightarrow \left( {x + 3} \right)\left( {{x^2} - 2x} \right) = 0\\
 \Leftrightarrow \left( {x + 3} \right).x.\left( {x - 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 3 = 0\\
x = 0\\
x - 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x =  - 3\\
x = 0\\
x = 2
\end{array} \right.\\
d,\,\\
{x^3} - 3{x^2} - 4x + 12 = 0\\
 \Leftrightarrow \left( {{x^3} - 3{x^2}} \right) - \left( {4x - 12} \right) = 0\\
 \Leftrightarrow {x^2}\left( {x - 3} \right) - 4.\left( {x - 3} \right) = 0\\
 \Leftrightarrow \left( {x - 3} \right)\left( {{x^2} - 4} \right) = 0\\
 \Leftrightarrow \left( {x - 3} \right)\left( {x - 2} \right)\left( {x + 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 3 = 0\\
x - 2 = 0\\
x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = 2\\
x =  - 2
\end{array} \right.
\end{array}\)