Đáp án:

\(\begin{array}{l}
1,\\
a,\,\,2x - 11\\
b,\,\,2x + 15\\
c,\,\,{x^2} + 3\\
2,\\
a,\,\,4\\
b,\,\,\, - 9\\
3,\\
a,\,\,\,\left( {x - y} \right)\left( {7x - 4} \right)\\
b,\,\,\,\left( {x - y + 3} \right)\left( {x + y + 3} \right)\\
c,\,\,\left( {x - 1} \right){\left( {x - 2} \right)^2}\\
d,\,\,\left( {x - y} \right)\left( {2 - x - y} \right)\\
e,\,\,\left( {x + 2y} \right)\left( {x - 2y - 2} \right)\\
f,\,\,\,x.\left( {x - y + 5} \right)\left( {x + y + 5} \right)\\
g,\,\,\,\left( {x - 3} \right)\left( {{x^2} - x + 9} \right)\\
h,\,\,\left( {x - 2} \right)\left( {x + 3} \right)\\
i,\,\,2.\left( {x - 3} \right)\left( {x + 5} \right)\\
4,\\
a,\,\,\left[ \begin{array}{l}
x = 0\\
x = 8\\
x =  - 8
\end{array} \right.\\
b,\,\,\,\left[ \begin{array}{l}
x = 5\\
x = \dfrac{1}{6}
\end{array} \right.\\
c,\,\,\left[ \begin{array}{l}
x =  - 1\\
x = 3\\
x =  - 2
\end{array} \right.\\
d,\,\,\left[ \begin{array}{l}
x = 0\\
x = 2
\end{array} \right.\\
e,\,\,\,x = 2\\
f,\,\,\left[ \begin{array}{l}
x = 4\\
x =  - 3
\end{array} \right.\\
g,\,\,\,\left[ \begin{array}{l}
x = 2\\
x =  - \dfrac{4}{3}
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
a,\,\,\left( {x - 3} \right)\left( {x + 5} \right) - \left( {x - 2} \right)\left( {x + 2} \right)\\
 = \left( {{x^2} + 5x - 3x - 15} \right) - \left( {{x^2} - {2^2}} \right)\\
 = \left( {{x^2} + 2x - 15} \right) - \left( {{x^2} - 4} \right)\\
 = {x^2} + 2x - 15 - {x^2} + 4\\
 = 2x - 11\\
b,\,\,{\left( {x - 2} \right)^2} + {\left( {x + 3} \right)^2} - 2.\left( {x - 1} \right)\left( {x + 1} \right)\\
 = \left( {{x^2} - 2.x.2 + {2^2}} \right) + \left( {{x^2} + 2.x.3 + {3^2}} \right) - 2.\left( {{x^2} - {1^2}} \right)\\
 = \left( {{x^2} - 4x + 4} \right) + \left( {{x^2} + 6x + 9} \right) - 2.\left( {{x^2} - 1} \right)\\
 = {x^2} - 4x + 4 + {x^2} + 6x + 9 - 2{x^2} + 2\\
 = \left( {{x^2} + {x^2} - 2{x^2}} \right) + \left( { - 4x + 6x} \right) + \left( {4 + 9 + 2} \right)\\
 = 2x + 15\\
c,\,\,{\left( {x - 2} \right)^2} + \left( {x - 1} \right)\left( {{x^2} + x + 1} \right) - x\left( {x - 2} \right)\left( {x + 2} \right)\\
 = \left( {{x^2} - 2.x.2 + {2^2}} \right) + \left( {x - 1} \right)\left( {{x^2} + x.1 + {1^2}} \right) - x.\left( {{x^2} - {2^2}} \right)\\
 = \left( {{x^2} - 4x + 4} \right) + \left( {{x^3} - {1^3}} \right) - x.\left( {{x^2} - 4} \right)\\
 = \left( {{x^2} - 4x + 4} \right) + \left( {{x^3} - 1} \right) - \left( {{x^3} - 4x} \right)\\
 = {x^2} - 4x + 4 + {x^3} - 1 - {x^3} + 4x\\
 = \left( {{x^3} - {x^3}} \right) + {x^2} + \left( { - 4x + 4x} \right) + \left( {4 - 1} \right)\\
 = {x^2} + 3\\
2,\\
a,\,\,{\left( {x - 1} \right)^2} - 2\left( {x - 3} \right)\left( {x - 1} \right) + {\left( {x - 3} \right)^2}\\
 = {\left( {x - 1} \right)^2} - 2.\left( {x - 1} \right).\left( {x - 3} \right) + {\left( {x - 3} \right)^2}\\
 = {\left[ {\left( {x - 1} \right) - \left( {x - 3} \right)} \right]^2}\\
 = {\left( {x - 1 - x + 3} \right)^2}\\
 = {2^2}\\
 = 4\\
b,\,\,\,{\left( {x - 1} \right)^3} - \left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right) + 3{x^2} - 3x\\
 = \left( {{x^3} - 3.{x^2}.1 + 3.x{{.1}^2} - {1^3}} \right) - \left( {x + 2} \right).\left( {{x^2} - x.2 + {2^2}} \right) + 3{x^2} - 3x\\
 = \left( {{x^3} - 3{x^2} + 3x - 1} \right) - \left( {{x^3} + {2^3}} \right) + 3{x^2} - 3x\\
 = {x^3} - 3{x^2} + 3x - 1 - {x^3} - 8 + 3{x^2} - 3x\\
 = \left( {{x^3} - {x^3}} \right) + \left( { - 3{x^2} + 3{x^2}} \right) + \left( {3x - 3x} \right) + \left( { - 1 - 8} \right)\\
 =  - 9\\
3,\\
a,\,\,\,7{x^2} - 7xy - 4x + 4y\\
 = \left( {7{x^2} - 7xy} \right) + \left( { - 4x + 4y} \right)\\
 = 7x.\left( {x - y} \right) - 4.\left( {x - y} \right)\\
 = \left( {x - y} \right)\left( {7x - 4} \right)\\
b,\,\,\,{x^2} + 6x - {y^2} + 9\\
 = \left( {{x^2} + 6x + 9} \right) - {y^2}\\
 = \left( {{x^2} + 2.x.3 + {3^2}} \right) - {y^2}\\
 = {\left( {x + 3} \right)^2} - {y^2}\\
 = \left[ {\left( {x + 3} \right) - y} \right]\left[ {\left( {x + 3} \right) + y} \right]\\
 = \left( {x - y + 3} \right)\left( {x + y + 3} \right)\\
c,\,\,{x^3} - {x^2} - 4{x^2} + 8x - 4\\
 = \left( {{x^3} - {x^2}} \right) - \left( {4{x^2} - 8x + 4} \right)\\
 = {x^2}\left( {x - 1} \right) - 4.\left( {{x^2} - 2x + 1} \right)\\
 = {x^2}\left( {x - 1} \right) - 4.\left( {{x^2} - 2.x.1 + {1^2}} \right)\\
 = {x^2}\left( {x - 1} \right) - 4.{\left( {x - 1} \right)^2}\\
 = \left( {x - 1} \right).\left[ {{x^2} - 4.\left( {x - 1} \right)} \right]\\
 = \left( {x - 1} \right)\left( {{x^2} - 4x + 4} \right)\\
 = \left( {x - 1} \right)\left( {{x^2} - 2.x.2 + {2^2}} \right)\\
 = \left( {x - 1} \right){\left( {x - 2} \right)^2}\\
d,\,\,2x - 2y - {x^2} + {y^2}\\
 = \left( {2x - 2y} \right) - \left( {{x^2} - {y^2}} \right)\\
 = 2.\left( {x - y} \right) - \left( {x - y} \right)\left( {x + y} \right)\\
 = \left( {x - y} \right).\left[ {2 - \left( {x + y} \right)} \right]\\
 = \left( {x - y} \right)\left( {2 - x - y} \right)\\
e,\,\,{x^2} - 2x - 4{y^2} - 4y\\
 = \left( {{x^2} - 4{y^2}} \right) + \left( { - 2x - 4y} \right)\\
 = \left[ {{x^2} - {{\left( {2y} \right)}^2}} \right] - 2.\left( {x + 2y} \right)\\
 = \left( {x - 2y} \right)\left( {x + 2y} \right) - 2\left( {x + 2y} \right)\\
 = \left( {x + 2y} \right).\left[ {\left( {x - 2y} \right) - 2} \right]\\
 = \left( {x + 2y} \right)\left( {x - 2y - 2} \right)\\
f,\,\,\,{x^3} + 10{x^2} + 25x - x{y^2}\\
 = x.\left( {{x^2} + 10x + 25 - {y^2}} \right)\\
 = x.\left[ {\left( {{x^2} + 10x + 25} \right) - {y^2}} \right]\\
 = x.\left[ {\left( {{x^2} + 2.x.5 + {5^2}} \right) - {y^2}} \right]\\
 = x.\left[ {{{\left( {x + 5} \right)}^2} - {y^2}} \right]\\
 = x.\left[ {\left( {x + 5} \right) - y} \right].\left[ {\left( {x + 5} \right) + y} \right]\\
 = x.\left( {x - y + 5} \right)\left( {x + y + 5} \right)\\
g,\,\,\,{x^3} - 4{x^2} + 12x - 27\\
 = \left( {{x^3} - 3{x^2}} \right) + \left( { - {x^2} + 3x} \right) + \left( {9x - 27} \right)\\
 = {x^2}\left( {x - 3} \right) - x.\left( {x - 3} \right) + 9\left( {x - 3} \right)\\
 = \left( {x - 3} \right)\left( {{x^2} - x + 9} \right)\\
h,\,\,{x^2} + x - 6\\
 = \left( {{x^2} - 2x} \right) + \left( {3x - 6} \right)\\
 = x.\left( {x - 2} \right) + 3.\left( {x - 2} \right)\\
 = \left( {x - 2} \right)\left( {x + 3} \right)\\
i,\,\,\,2{x^2} + 4x - 30\\
 = 2.\left( {{x^2} + 2x - 15} \right)\\
 = 2.\left[ {\left( {{x^2} - 3x} \right) + \left( {5x - 15} \right)} \right]\\
 = 2.\left[ {x\left( {x - 3} \right) + 5.\left( {x - 3} \right)} \right]\\
 = 2.\left( {x - 3} \right)\left( {x + 5} \right)\\
4,\\
a,\,\,{x^3} - 64x = 0\\
 \Leftrightarrow x.\left( {{x^2} - 64} \right) = 0\\
 \Leftrightarrow x\left( {{x^2} - {8^2}} \right) = 0\\
 \Leftrightarrow x\left( {x - 8} \right)\left( {x + 8} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x - 8 = 0\\
x + 8 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 8\\
x =  - 8
\end{array} \right.\\
b,\,\,\,6x\left( {x - 5} \right) = x - 5\\
 \Leftrightarrow 6x\left( {x - 5} \right) - \left( {x - 5} \right) = 0\\
 \Leftrightarrow \left( {x - 5} \right)\left( {6x - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 5 = 0\\
6x - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 5\\
x = \dfrac{1}{6}
\end{array} \right.\\
c,\,\,{x^3} - 7x - 6 = 0\\
 \Leftrightarrow \left( {{x^3} + {x^2}} \right) + \left( { - {x^2} - x} \right) + \left( { - 6x - 6} \right) = 0\\
 \Leftrightarrow {x^2}\left( {x + 1} \right) - x.\left( {x + 1} \right) - 6.\left( {x + 1} \right) = 0\\
 \Leftrightarrow \left( {x + 1} \right)\left( {{x^2} - x - 6} \right) = 0\\
 \Leftrightarrow \left( {x + 1} \right).\left[ {\left( {{x^2} - 3x} \right) + \left( {2x - 6} \right)} \right] = 0\\
 \Leftrightarrow \left( {x + 1} \right).\left[ {x\left( {x - 3} \right) + 2.\left( {x - 3} \right)} \right] = 0\\
 \Leftrightarrow \left( {x + 1} \right)\left( {x - 3} \right)\left( {x + 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 1 = 0\\
x - 3 = 0\\
x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x =  - 1\\
x = 3\\
x =  - 2
\end{array} \right.\\
d,\,\,{x^3} - 4{x^2} =  - 4x\\
 \Leftrightarrow {x^3} - 4{x^2} + 4x = 0\\
 \Leftrightarrow x\left( {{x^2} - 4x + 4} \right) = 0\\
 \Leftrightarrow x.\left( {{x^2} - 2.x.2 + {2^2}} \right) = 0\\
 \Leftrightarrow x.{\left( {x - 2} \right)^2} = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
{\left( {x - 2} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 2
\end{array} \right.\\
e,\,\,\,{x^3} - 6{x^2} + 12x - 8 = 0\\
 \Leftrightarrow {x^3} - 3.{x^2}.2 + 3.x{.2^2} - {2^3} = 0\\
 \Leftrightarrow {\left( {x - 2} \right)^3} = 0\\
 \Leftrightarrow x - 2 = 0\\
 \Leftrightarrow x = 2\\
f,\,\,{x^2} - 16 - \left( {x - 4} \right) = 0\\
 \Leftrightarrow {x^2} - {4^2} - \left( {x - 4} \right) = 0\\
 \Leftrightarrow \left( {x - 4} \right)\left( {x + 4} \right) - \left( {x - 4} \right) = 0\\
 \Leftrightarrow \left( {x - 4} \right).\left[ {\left( {x + 4} \right) - 1} \right] = 0\\
 \Leftrightarrow \left( {x - 4} \right)\left( {x + 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 4 = 0\\
x + 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 4\\
x =  - 3
\end{array} \right.\\
g,\,\,\,{\left( {2x + 1} \right)^2} = {\left( {3 + x} \right)^2}\\
 \Leftrightarrow {\left( {2x + 1} \right)^2} - {\left( {x + 3} \right)^2} = 0\\
 \Leftrightarrow \left[ {\left( {2x + 1} \right) - \left( {x + 3} \right)} \right].\left[ {\left( {2x + 1} \right) + \left( {x + 3} \right)} \right] = 0\\
 \Leftrightarrow \left( {2x + 1 - x - 3} \right)\left( {2x + 1 + x + 3} \right) = 0\\
 \Leftrightarrow \left( {x - 2} \right)\left( {3x + 4} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
3x + 4 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x =  - \dfrac{4}{3}
\end{array} \right.
\end{array}\)