Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
2,\\
a,\\
3x.\left( {x - 1} \right) - x + 1 = 0\\
 \Leftrightarrow 3x.\left( {x - 1} \right) - \left( {x - 1} \right) = 0\\
 \Leftrightarrow \left( {x - 1} \right)\left( {3x - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
3x - 1 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = \dfrac{1}{3}
\end{array} \right.\\
b,\\
{x^3} - 8 + \left( {x - 2} \right)\left( {x + 7} \right) = 0\\
 \Leftrightarrow {x^3} - {2^3} + \left( {x - 2} \right)\left( {x + 7} \right) = 0\\
 \Leftrightarrow \left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right) + \left( {x - 2} \right)\left( {x + 7} \right) = 0\\
 \Leftrightarrow \left( {x - 2} \right).\left[ {\left( {{x^2} + 2x + 4} \right) + \left( {x + 7} \right)} \right] = 0\\
 \Leftrightarrow \left( {x - 2} \right).\left( {{x^2} + 3x + 11} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
{x^2} + 3x + 11 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
\left( {{x^2} + 3x + \dfrac{9}{4}} \right) + \dfrac{{35}}{4} = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
{\left( {x + \dfrac{3}{2}} \right)^2} + \dfrac{{35}}{4} = 0\,\,\,\,\,\left( {vn} \right)
\end{array} \right.\\
 \Leftrightarrow x = 2\\
c,\\
{x^3} + 2{x^2} - x - 2 = 0\\
 \Leftrightarrow \left( {{x^3} + 2{x^2}} \right) - \left( {x + 2} \right) = 0\\
 \Leftrightarrow {x^2}\left( {x + 2} \right) - \left( {x + 2} \right) = 0\\
 \Leftrightarrow \left( {x + 2} \right)\left( {{x^2} - 1} \right) = 0\\
 \Leftrightarrow \left( {x + 2} \right)\left( {x - 1} \right)\left( {x + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 2 = 0\\
x - 1 = 0\\
x + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x =  - 2\\
x = 1\\
x =  - 1
\end{array} \right.\\
3,\\
a,\\
A = {x^2} - 4x + 8\\
 = \left( {{x^2} - 4x + 4} \right) + 4\\
 = {\left( {x - 2} \right)^2} + 4 \ge 4,\,\,\,\forall x\\
 \Rightarrow {A_{\min }} = 4 \Leftrightarrow {\left( {x - 2} \right)^2} = 0 \Leftrightarrow x = 2\\
b,\\
B = 6 - 2x - {x^2}\\
 = 7 + \left( { - 1 - 2x - {x^2}} \right)\\
 = 7 - \left( {{x^2} + 2x + 1} \right)\\
 = 7 - {\left( {x + 1} \right)^2} \le 7,\,\,\,\forall x\\
 \Rightarrow {B_{\max }} = 7 \Leftrightarrow {\left( {x + 1} \right)^2} = 0 \Leftrightarrow x =  - 1\\
c,\\
C = {x^2} + 3x + 3\\
 = \left( {{x^2} + 3x + \dfrac{9}{4}} \right) + \dfrac{3}{4}\\
 = \left( {{x^2} + 2.x.\dfrac{3}{2} + {{\left( {\dfrac{3}{2}} \right)}^2}} \right) + \dfrac{3}{4}\\
 = {\left( {x + \dfrac{3}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4},\,\,\,\forall x\\
 \Rightarrow {C_{\min }} = \dfrac{3}{4} \Leftrightarrow {\left( {x + \dfrac{3}{2}} \right)^2} = 0 \Leftrightarrow x =  - \dfrac{3}{2}
\end{array}\)