Đáp án:

\(\begin{array}{l}
a,\\
x = \dfrac{1}{2}\\
b,\\
\left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.\\
c,\\
x = \dfrac{3}{2}\\
d,\\
x = 4\\
e,\\
\left[ \begin{array}{l}
x = 1\\
x = 6
\end{array} \right.\\
g,\\
\left[ \begin{array}{l}
x =  - 1\\
x = \dfrac{7}{2}
\end{array} \right.\\
h,\\
 \Leftrightarrow x =  - 1
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
\left( {6x - 1} \right)\left( {x - 2} \right) - \left( {2x - 3} \right)\left( {3x + 1} \right) = 2\\
 \Leftrightarrow \left( {6{x^2} - 12x - x + 2} \right) - \left( {6{x^2} + 2x - 9x - 3} \right) = 2\\
 \Leftrightarrow \left( {6{x^2} - 13x + 2} \right) - \left( {6{x^2} - 7x - 3} \right) = 2\\
 \Leftrightarrow 6{x^2} - 13x + 2 - 6{x^2} + 7x + 3 = 2\\
 \Leftrightarrow \left( {6{x^2} - 6{x^2}} \right) + \left( { - 13x + 7x} \right) + \left( {2 + 3} \right) = 2\\
 \Leftrightarrow  - 6x + 5 = 2\\
 \Leftrightarrow  - 6x =  - 3\\
 \Leftrightarrow x = \dfrac{1}{2}\\
b,\\
{x^4} - 2{x^3} + {x^2} = 0\\
 \Leftrightarrow {x^2}.\left( {{x^2} - 2x + 1} \right) = 0\\
 \Leftrightarrow {x^2}.\left( {{x^2} - 2.x.1 + {1^2}} \right) = 0\\
 \Leftrightarrow {x^2}.{\left( {x - 1} \right)^2} = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
{x^2} = 0\\
{\left( {x - 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.\\
c,\\
{\left( {x - 1} \right)^2} - {\left( {x - 2} \right)^2} = 0\\
 \Leftrightarrow \left[ {\left( {x - 1} \right) - \left( {x - 2} \right)} \right].\left[ {\left( {x - 1} \right) + \left( {x - 2} \right)} \right] = 0\\
 \Leftrightarrow \left( {x - 1 - x + 2} \right).\left( {x - 1 + x - 2} \right) = 0\\
 \Leftrightarrow 1.\left( {2x - 3} \right) = 0\\
 \Leftrightarrow 2x - 3 = 0\\
 \Leftrightarrow x = \dfrac{3}{2}\\
d,\\
8 - {\left( {x - 2} \right)^3} = 0\\
 \Leftrightarrow {\left( {x - 2} \right)^3} = 8\\
 \Leftrightarrow {\left( {x - 2} \right)^3} = {2^3}\\
 \Leftrightarrow x - 2 = 2\\
 \Leftrightarrow x = 4\\
e,\\
{x^2} - 7x + 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.\\
g,\\
2{x^2} - 5x - 7 = 0\\
 \Leftrightarrow \left( {2{x^2} + 2x} \right) + \left( { - 7x - 7} \right) = 0\\
 \Leftrightarrow 2x.\left( {x + 1} \right) - 7.\left( {x + 1} \right) = 0\\
 \Leftrightarrow \left( {x + 1} \right)\left( {2x - 7} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 1 = 0\\
2x - 7 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - 1\\
x = \dfrac{7}{2}
\end{array} \right.\\
h,\\
{x^4} + 2{x^3} + 2{x^2} + 2x + 1 = 0\\
 \Leftrightarrow \left( {{x^4} + 2{x^3} + {x^2}} \right) + \left( {{x^2} + 2x + 1} \right) = 0\\
 \Leftrightarrow {x^2}.\left( {{x^2} + 2x + 1} \right) + \left( {{x^2} + 2x + 1} \right) = 0\\
 \Leftrightarrow \left( {{x^2} + 2x + 1} \right)\left( {{x^2} + 1} \right) = 0\\
 \Leftrightarrow \left( {{x^2} + 2.x.1 + {1^2}} \right)\left( {{x^2} + 1} \right) = 0\\
 \Leftrightarrow {\left( {x + 1} \right)^2}.\left( {{x^2} + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
{\left( {x + 1} \right)^2} = 0\\
{x^2} + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x + 1 = 0\\
{x^2} =  - 1\,\,\,\left( L \right)
\end{array} \right. \Leftrightarrow x =  - 1
\end{array}\)