Đáp án:

\(\begin{array}{l}
9,\,\,\,\,D\\
10,\,\,\,\,A\\
11,\,\,\,\,D\\
12,\,\,\,\,D\\
13,\,\,\,\,\,C\\
14,\,\,\,\,B\\
15,\,\,\,\,\,C\\
16,\,\,\,\,\,A
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
9,\\
x.\left( {5 - 10x} \right) - 3.\left( {10x - 5} \right) = 0\\
 \Leftrightarrow x.\left( {5 - 10x} \right) - 3.\left[ { - \left( {5 - 10x} \right)} \right] = 0\\
 \Leftrightarrow x.\left( {5 - 10x} \right) + 3.\left( {5 - 10x} \right) = 0\\
 \Leftrightarrow \left( {5 - 10x} \right).\left( {x + 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
5 - 10x = 0\\
x + 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
10x = 5\\
x =  - 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x =  - 3
\end{array} \right.\\
 \Rightarrow {x_1} = \dfrac{1}{2};\,\,\,{x_2} =  - 3\\
 \Rightarrow {x_1} + {x_2} = \dfrac{1}{2} + \left( { - 3} \right) =  - \dfrac{7}{2}\\
10,\\
{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)\\
11,\\
25 - {a^2} + 2ab - {b^2}\\
 = 25 - \left( {{a^2} - 2ab + {b^2}} \right)\\
 = {5^2} - {\left( {a - b} \right)^2}\\
 = \left[ {5 - \left( {a - b} \right)} \right].\left[ {5 + \left( {a - b} \right)} \right]\\
 = \left( {5 - a + b} \right)\left( {5 + a - b} \right)\\
12,\\
m.{n^3} - 1 + m - {n^3}\\
 = \left( {m.{n^3} + m} \right) + \left( { - {n^3} - 1} \right)\\
 = m.\left( {{n^3} + 1} \right) - \left( {{n^3} + 1} \right)\\
 = \left( {{n^3} + 1} \right)\left( {m - 1} \right)\\
13,\\
{x^4} + 64\\
 = \left( {{x^4} + 16{x^2} + 64} \right) - 16{x^2}\\
 = \left[ {{{\left( {{x^2}} \right)}^2} + 2.{x^2}.8 + {8^2}} \right] - 16{x^2}\\
 = {\left( {{x^2} + 8} \right)^2} - {\left( {4x} \right)^2}\\
14,\\
4{x^2} + 4x - {y^2} + 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)\\
 = \left( {2x - y + 1} \right)\left( {2x + y + 1} \right)\\
15,\\
*)\\
3{x^2} - 5x - 2 = \left( {3{x^2} - 6x} \right) + \left( {x - 2} \right)\\
 = 3x.\left( {x - 2} \right) + \left( {x - 2} \right) = \left( {x - 2} \right)\left( {3x + 1} \right)\\
*)\\
{x^2} + 5x + 4 = \left( {{x^2} + 4x} \right) + \left( {x + 4} \right)\\
 = x\left( {x + 4} \right) + \left( {x + 4} \right) = \left( {x + 4} \right)\left( {x + 1} \right)\\
*)\\
{x^2} - 9x + 8 = \left( {{x^2} - x} \right) + \left( { - 8x + 8} \right)\\
 = x\left( {x - 1} \right) - 8.\left( {x - 1} \right) = \left( {x - 1} \right)\left( {x - 8} \right)\\
*)\\
{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)\\
16,\\
*)\\
{x^4} + 4{x^2} - 5 = \left( {{x^4} - {x^2}} \right) + \left( {5{x^2} - 5} \right)\\
 = {x^2}.\left( {{x^2} - 1} \right) + 5.\left( {{x^2} - 1} \right) = \left( {{x^2} - 1} \right)\left( {{x^2} + 5} \right)\\
 = \left( {{x^2} - {1^2}} \right)\left( {{x^2} + 5} \right) = \left( {x - 1} \right)\left( {x + 1} \right)\left( {{x^2} + 5} \right)\\
*)\\
{x^2} + 5x + 4 = \left( {{x^2} + x} \right) + \left( {4x + 4} \right)\\
 = x\left( {x + 1} \right) + 4.\left( {x + 1} \right) = \left( {x + 1} \right)\left( {x + 4} \right)\\
*)\\
{x^2} - 9x + 8 = \left( {{x^2} - x} \right) + \left( { - 8x + 8} \right)\\
 = x\left( {x - 1} \right) - 8.\left( {x - 1} \right) = \left( {x - 1} \right)\left( {x - 8} \right)\\
*)\\
{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)
\end{array}\)

\(\begin{array}{l}
9,\\
x.\left( {5 - 10x} \right) - 3.\left( {10x - 5} \right) = 0\\
 \Leftrightarrow x.\left( {5 - 10x} \right) - 3.\left[ { - \left( {5 - 10x} \right)} \right] = 0\\
 \Leftrightarrow x.\left( {5 - 10x} \right) + 3.\left( {5 - 10x} \right) = 0\\
 \Leftrightarrow \left( {5 - 10x} \right).\left( {x + 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
5 - 10x = 0\\
x + 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
10x = 5\\
x =  - 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x =  - 3
\end{array} \right.\\
 \Rightarrow {x_1} = \dfrac{1}{2};\,\,\,{x_2} =  - 3\\
 \Rightarrow {x_1} + {x_2} = \dfrac{1}{2} + \left( { - 3} \right) =  - \dfrac{7}{2}\\
10,\\
{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)\\
11,\\
25 - {a^2} + 2ab - {b^2}\\
 = 25 - \left( {{a^2} - 2ab + {b^2}} \right)\\
 = {5^2} - {\left( {a - b} \right)^2}\\
 = \left[ {5 - \left( {a - b} \right)} \right].\left[ {5 + \left( {a - b} \right)} \right]\\
 = \left( {5 - a + b} \right)\left( {5 + a - b} \right)\\
12,\\
m.{n^3} - 1 + m - {n^3}\\
 = \left( {m.{n^3} + m} \right) + \left( { - {n^3} - 1} \right)\\
 = m.\left( {{n^3} + 1} \right) - \left( {{n^3} + 1} \right)\\
 = \left( {{n^3} + 1} \right)\left( {m - 1} \right)\\
13,\\
{x^4} + 64\\
 = \left( {{x^4} + 16{x^2} + 64} \right) - 16{x^2}\\
 = \left[ {{{\left( {{x^2}} \right)}^2} + 2.{x^2}.8 + {8^2}} \right] - 16{x^2}\\
 = {\left( {{x^2} + 8} \right)^2} - {\left( {4x} \right)^2}\\
14,\\
4{x^2} + 4x - {y^2} + 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)\\
 = \left( {2x - y + 1} \right)\left( {2x + y + 1} \right)\\
15,\\
*)\\
3{x^2} - 5x - 2 = \left( {3{x^2} - 6x} \right) + \left( {x - 2} \right)\\
 = 3x.\left( {x - 2} \right) + \left( {x - 2} \right) = \left( {x - 2} \right)\left( {3x + 1} \right)\\
*)\\
{x^2} + 5x + 4 = \left( {{x^2} + 4x} \right) + \left( {x + 4} \right)\\
 = x\left( {x + 4} \right) + \left( {x + 4} \right) = \left( {x + 4} \right)\left( {x + 1} \right)\\
*)\\
{x^2} - 9x + 8 = \left( {{x^2} - x} \right) + \left( { - 8x + 8} \right)\\
 = x\left( {x - 1} \right) - 8.\left( {x - 1} \right) = \left( {x - 1} \right)\left( {x - 8} \right)\\
*)\\
{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)\\
16,\\
*)\\
{x^4} + 4{x^2} - 5 = \left( {{x^4} - {x^2}} \right) + \left( {5{x^2} - 5} \right)\\
 = {x^2}.\left( {{x^2} - 1} \right) + 5.\left( {{x^2} - 1} \right) = \left( {{x^2} - 1} \right)\left( {{x^2} + 5} \right)\\
 = \left( {{x^2} - {1^2}} \right)\left( {{x^2} + 5} \right) = \left( {x - 1} \right)\left( {x + 1} \right)\left( {{x^2} + 5} \right)\\
*)\\
{x^2} + 5x + 4 = \left( {{x^2} + x} \right) + \left( {4x + 4} \right)\\
 = x\left( {x + 1} \right) + 4.\left( {x + 1} \right) = \left( {x + 1} \right)\left( {x + 4} \right)\\
*)\\
{x^2} - 9x + 8 = \left( {{x^2} - x} \right) + \left( { - 8x + 8} \right)\\
 = x\left( {x - 1} \right) - 8.\left( {x - 1} \right) = \left( {x - 1} \right)\left( {x - 8} \right)\\
*)\\
{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)
\end{array}\)