Đáp án:

\(\begin{array}{l}
1,\\
a,\,\,\, - x + 15\\
b,\,\,\,2\\
c,\,\,\, - 4x - 10\\
d,\,\,\,x - 3\\
2,\\
a,\,\,\left( {x + y} \right)\left( {1 - y} \right)\\
b,\,\,\,\left( {x - y - 2} \right)\left( {x + y - 2} \right)\\
c,\,\,\,\left( {x - 3} \right)\left( {x + 1} \right)\\
d,\,\,\,\left( {x - 4} \right)\left( {x + 1} \right).\left( {x - 5} \right)\left( {x + 2} \right)\\
3,\\
a,\,\,\,\left[ \begin{array}{l}
x = 0\\
x =  - 3
\end{array} \right.\\
b,\,\,\,\left[ \begin{array}{l}
x = 0\\
x = 2\\
x =  - 2
\end{array} \right.\\
c,\,\,\,\left[ \begin{array}{l}
x = 1\\
x =  - 6
\end{array} \right.\\
d,\,\,\,\left[ \begin{array}{l}
x = 1\\
x = 2014
\end{array} \right.\\
4,\\
a,\,\,\,a = 30\\
b,\,\,\,\,n =  - 12\\
5,\\
a,\,\,\,\,x - {x^2} - 1 < 0,\,\,\,\forall x\\
b,\,\,\,\,f{\left( x \right)_{\min }} = 5 \Leftrightarrow x = 2
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
a,\,\,\,\left( {4x - 3} \right)\left( {x - 5} \right) - 2x\left( {2x - 11} \right)\\
 = 4x\left( {x - 5} \right) - 3\left( {x - 5} \right) - \left( {2x.2x - 2x.11} \right)\\
 = \left( {4x.x - 4x.5} \right) - \left( {3.x - 3.5} \right) - \left( {4{x^2} - 22x} \right)\\
 = \left( {4{x^2} - 20x} \right) - \left( {3x - 15} \right) - 4{x^2} + 22x\\
 = 4{x^2} - 20x - 3x + 15 - 4{x^2} + 22x\\
 = \left( {4{x^2} - 4{x^2}} \right) + \left( { - 20x - 3x + 22x} \right) + 15\\
 =  - x + 15\\
b,\,\,\,\left( {x + 1} \right)\left( {{x^2} - x + 1} \right) - \left( {x - 1} \right)\left( {{x^2} + x + 1} \right)\\
 = \left( {x + 1} \right)\left( {{x^2} - x.1 + {1^2}} \right) - \left( {x - 1} \right)\left( {{x^2} + x.1 + {1^2}} \right)\\
 = \left( {{x^3} + {1^3}} \right) - \left( {{x^3} - {1^3}} \right)\\
 = \left( {{x^3} + 1} \right) - \left( {{x^3} - 1} \right)\\
 = {x^3} + 1 - {x^3} + 1\\
 = 2\\
c,\,\,\,\left( {2x + 3} \right)\left( {2x - 3} \right) - {\left( {2x + 1} \right)^2}\\
 = \left[ {{{\left( {2x} \right)}^2} - {3^2}} \right] - \left[ {{{\left( {2x} \right)}^2} + 2.2x.1 + {1^2}} \right]\\
 = \left( {4{x^2} - 9} \right) - \left( {4{x^2} + 4x + 1} \right)\\
 = 4{x^2} - 9 - 4{x^2} - 4x - 1\\
 = \left( {4{x^2} - 4{x^2}} \right) - 4x + \left( { - 9 - 1} \right)\\
 =  - 4x - 10\\
d,\,\,\,\left( {{x^2} - 3x + xy - 3y} \right):\left( {x + y} \right)\\
 = \left[ {\left( {{x^2} - 3x} \right) + \left( {xy - 3y} \right)} \right]:\left( {x + y} \right)\\
 = \left[ {x\left( {x - 3} \right) + y\left( {x - 3} \right)} \right]:\left( {x + y} \right)\\
 = \left[ {\left( {x + y} \right)\left( {x - 3} \right)} \right]:\left( {x + y} \right)\\
 = x - 3\\
2,\\
a,\,\,x - xy + y - {y^2}\\
 = \left( {x + y} \right) - \left( {xy + {y^2}} \right)\\
 = \left( {x + y} \right) - y.\left( {x + y} \right)\\
 = \left( {x + y} \right)\left( {1 - y} \right)\\
b,\,\,\,{x^2} - 4x - {y^2} + 4\\
 = \left( {{x^2} - 4x + 4} \right) - {y^2}\\
 = \left( {{x^2} - 2.x.2 + {2^2}} \right) - {y^2}\\
 = {\left( {x - 2} \right)^2} - {y^2}\\
 = \left[ {\left( {x - 2} \right) - y} \right].\left[ {\left( {x - 2} \right) + y} \right]\\
 = \left( {x - y - 2} \right)\left( {x + y - 2} \right)\\
c,\,\,\,{x^2} - 2x - 3\\
 = \left( {{x^2} - 3x} \right) + \left( {x - 3} \right)\\
 = x\left( {x - 3} \right) + \left( {x - 3} \right)\\
 = \left( {x - 3} \right)\left( {x + 1} \right)\\
d,\,\,\,{\left( {{x^2} - 3x - 1} \right)^2} - 12\left( {{x^2} - 3x - 1} \right) + 27\\
 = \left[ {{{\left( {{x^2} - 3x - 1} \right)}^2} - 3\left( {{x^2} - 3x - 1} \right)} \right] + \left[ { - 9\left( {{x^2} - 3x - 1} \right) + 27} \right]\\
 = \left( {{x^2} - 3x - 1} \right).\left[ {\left( {{x^2} - 3x - 1} \right) - 3} \right] - 9\left[ {\left( {{x^2} - 3x - 1} \right) - 3} \right]\\
 = \left[ {\left( {{x^2} - 3x - 1} \right) - 3} \right].\left[ {\left( {{x^2} - 3x - 1} \right) - 9} \right]\\
 = \left( {{x^2} - 3x - 4} \right)\left( {{x^2} - 3x - 10} \right)\\
 = \left[ {\left( {{x^2} - 4x} \right) + \left( {x - 4} \right)} \right].\left[ {\left( {{x^2} - 5x} \right) + \left( {2x - 10} \right)} \right]\\
 = \left[ {x.\left( {x - 4} \right) + \left( {x - 4} \right)} \right].\left[ {x\left( {x - 5} \right) + 2\left( {x - 5} \right)} \right]\\
 = \left( {x - 4} \right)\left( {x + 1} \right).\left( {x - 5} \right)\left( {x + 2} \right)\\
3,\\
a,\,\,\,\,{x^2} + 3x = 0\\
 \Leftrightarrow x\left( {x + 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x + 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x =  - 3
\end{array} \right.\\
b,\,\,\,{x^3} - 4x = 0\\
 \Leftrightarrow x\left( {{x^2} - 4} \right) = 0\\
 \Leftrightarrow x\left( {{x^2} - {2^2}} \right) = 0\\
 \Leftrightarrow x\left( {x - 2} \right)\left( {x + 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x - 2 = 0\\
x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 2\\
x =  - 2
\end{array} \right.\\
c,\,\,\,{x^2} + 5x = 6\\
 \Leftrightarrow {x^2} + 5x - 6 = 0\\
 \Leftrightarrow \left( {{x^2} - x} \right) + \left( {6x - 6} \right) = 0\\
 \Leftrightarrow x\left( {x - 1} \right) + 6\left( {x - 1} \right) = 0\\
 \Leftrightarrow \left( {x - 1} \right)\left( {x + 6} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
x + 6 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x =  - 6
\end{array} \right.\\
d,\,\,\,{x^2} - 2015x + 2014 = 0\\
 \Leftrightarrow \left( {{x^2} - x} \right) + \left( { - 2014x + 2014} \right) = 0\\
 \Leftrightarrow x\left( {x - 1} \right) - 2014\left( {x - 1} \right) = 0\\
 \Leftrightarrow \left( {x - 1} \right)\left( {x - 2014} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
x - 2014 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = 2014
\end{array} \right.\\
4,\\
a,\,\,\\
f\left( x \right) = 2{x^3} - 3{x^2} + x + a\\
g\left( x \right) = x + 2\\
g\left( x \right) = 0 \Leftrightarrow x + 2 = 0 \Leftrightarrow x =  - 2\\
f\left( x \right) \vdots g\left( x \right) \Leftrightarrow f\left( { - 2} \right) = 0\\
 \Leftrightarrow 2.{\left( { - 2} \right)^3} - 3.{\left( { - 2} \right)^2} + \left( { - 2} \right) + a = 0\\
 \Leftrightarrow 2.\left( { - 8} \right) - 3.4 - 2 + a = 0\\
 \Leftrightarrow  - 30 + a = 0\\
 \Leftrightarrow a = 30\\
b,\\
f\left( x \right) = {x^2} + 4x + n\\
g\left( x \right) = x - 2\\
g\left( x \right) = 0 \Leftrightarrow x - 2 = 0 \Leftrightarrow x = 2\\
f\left( x \right) \vdots g\left( x \right) \Leftrightarrow f\left( 2 \right) = 0\\
 \Leftrightarrow {2^2} + 4.2 + n = 0\\
 \Leftrightarrow 4 + 8 + n = 0\\
 \Leftrightarrow 12 + n = 0\\
 \Leftrightarrow n =  - 12\\
5,\\
a,\\
A = x - {x^2} - 1 =  - \dfrac{3}{4} + \left( {x - {x^2} - \dfrac{1}{4}} \right)\\
 =  - \dfrac{3}{4} - \left( {{x^2} - x + \dfrac{1}{4}} \right)\\
 =  - \dfrac{3}{4} - \left[ {{x^2} - 2.x.\dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2}} \right]\\
 =  - \dfrac{3}{4} - {\left( {x - \dfrac{1}{2}} \right)^2}\\
{\left( {x - \dfrac{1}{2}} \right)^2} \ge 0,\,\,\,\forall x\\
 \Rightarrow A =  - \dfrac{3}{4} - {\left( {x - \dfrac{1}{2}} \right)^2} \le  - \dfrac{3}{4} < 0,\,\,\,\forall x\\
 \Rightarrow x - {x^2} - 1 < 0,\,\,\,\forall x\\
b,\\
f\left( x \right) = {x^2} - 4x + 9 = \left( {{x^2} - 4x + 4} \right) + 5\\
 = \left( {{x^2} - 2.x.2 + {2^2}} \right) + 5\\
 = {\left( {x - 2} \right)^2} + 5\\
{\left( {x - 2} \right)^2} \ge 0,\,\,\,\forall x\\
 \Rightarrow f\left( x \right) = {\left( {x - 2} \right)^2} + 5 \ge 5,\,\,\,\,\forall x\\
 \Rightarrow f{\left( x \right)_{\min }} = 5 \Leftrightarrow {\left( {x - 2} \right)^2} = 0 \Leftrightarrow x - 2 = 0 \Leftrightarrow x = 2\\
 \Rightarrow f{\left( x \right)_{\min }} = 5 \Leftrightarrow x = 2
\end{array}\)