Đáp án:

\(\begin{array}{l}
2,\\
a,\\
\left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = 1\\
x =  - 4
\end{array} \right.\\
b,\\
\left[ \begin{array}{l}
x = 0\\
x = 2
\end{array} \right.\\
c,\\
\left[ \begin{array}{l}
x = 1\\
x = 2\\
x = \dfrac{1}{2}
\end{array} \right.\\
d,\\
x = 2\\
3,\\
a,\\
\left[ \begin{array}{l}
x = 2\\
x =  - 1
\end{array} \right.\\
b,\\
\left[ \begin{array}{l}
x = 2\\
x = \dfrac{2}{3}
\end{array} \right.\\
c,\\
Phương\,\,trình\,\,vô\,\,nghiệm\\
d,\\
\left[ \begin{array}{l}
x = 1\\
x =  - \dfrac{6}{5}
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
2,\\
a,\\
\left( {2x - 1} \right)\left( {{x^2} + 3x - 4} \right) = 0\\
 \Leftrightarrow \left( {2x - 1} \right).\left[ {\left( {{x^2} - x} \right) + \left( {4x - 4} \right)} \right] = 0\\
 \Leftrightarrow \left( {2x - 1} \right)\left[ {x\left( {x - 1} \right) + 4\left( {x - 1} \right)} \right] = 0\\
 \Leftrightarrow \left( {2x - 1} \right)\left( {x - 1} \right)\left( {x + 4} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - 1 = 0\\
x - 1 = 0\\
x + 4 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = 1\\
x =  - 4
\end{array} \right.\\
b,\\
\left( {2 - x} \right)\left( {{x^2} - 2x} \right) = 0\\
 \Leftrightarrow \left( {2 - x} \right).x.\left( {x - 2} \right) = 0\\
 \Leftrightarrow  - \left( {x - 2} \right).x\left( {x - 2} \right) = 0\\
 \Leftrightarrow x{\left( {x - 2} \right)^2} = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x - 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 2
\end{array} \right.\\
c,\\
\left( {x - 1} \right)\left( {2{x^2} - 5x + 2} \right) = 0\\
 \Leftrightarrow \left( {x - 1} \right).\left[ {\left( {2{x^2} - 4x} \right) + \left( { - x + 2} \right)} \right] = 0\\
 \Leftrightarrow \left( {x - 1} \right).\left[ {2x.\left( {x - 2} \right) - \left( {x - 2} \right)} \right] = 0\\
 \Leftrightarrow \left( {x - 1} \right)\left( {x - 2} \right)\left( {2x - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
x - 2 = 0\\
2x - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = 2\\
x = \dfrac{1}{2}
\end{array} \right.\\
d,\\
\left( {2x - 3} \right)\left( {x - 2} \right) = x - 2\\
 \Leftrightarrow \left( {2x - 3} \right)\left( {x - 2} \right) - \left( {x - 2} \right) = 0\\
 \Leftrightarrow \left( {x - 2} \right).\left[ {\left( {2x - 3} \right) - 1} \right] = 0\\
 \Leftrightarrow \left( {x - 2} \right)\left( {2x - 4} \right) = 0\\
 \Leftrightarrow \left( {x - 2} \right).2.\left( {x - 2} \right) = 0\\
 \Leftrightarrow 2{\left( {x - 2} \right)^2} = 0\\
 \Leftrightarrow x - 2 = 0\\
 \Leftrightarrow x = 2\\
3,\\
a,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ne 1\\
x \ne  - 2
\end{array} \right.\\
\dfrac{2}{{x - 1}} + \dfrac{4}{{x + 2}} = 3\\
 \Leftrightarrow \dfrac{{2.\left( {x + 2} \right) + 4.\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 2} \right)}} = 3\\
 \Leftrightarrow \dfrac{{2x + 4 + 4x - 4}}{{{x^2} + 2x - x - 2}} = 3\\
 \Leftrightarrow \dfrac{{6x}}{{{x^2} + x - 2}} = 3\\
 \Leftrightarrow \dfrac{{2x}}{{{x^2} + x - 2}} = 1\\
 \Leftrightarrow 2x = {x^2} + x - 2\\
 \Leftrightarrow \left( {{x^2} + x - 2} \right) - 2x = 0\\
 \Leftrightarrow {x^2} - x - 2 = 0\\
 \Leftrightarrow \left( {{x^2} - 2x} \right) + \left( {x - 2} \right) = 0\\
 \Leftrightarrow x\left( {x - 2} \right) + \left( {x - 2} \right) = 0\\
 \Leftrightarrow \left( {x - 2} \right)\left( {x + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
x + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x =  - 1
\end{array} \right.\\
b,\\
DKXD:\,\,\,x \ne 1\\
\dfrac{x}{{x - 1}} = 3x - 4\\
 \Leftrightarrow x = \left( {x - 1} \right)\left( {3x - 4} \right)\\
 \Leftrightarrow x = 3{x^2} - 4x - 3x + 4\\
 \Leftrightarrow x = 3{x^2} - 7x + 4\\
 \Leftrightarrow 3{x^2} - 8x + 4 = 0\\
 \Leftrightarrow \left( {3{x^2} - 6x} \right) + \left( { - 2x + 4} \right) = 0\\
 \Leftrightarrow 3x.\left( {x - 2} \right) - 2.\left( {x - 2} \right) = 0\\
 \Leftrightarrow \left( {x - 2} \right)\left( {3x - 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
3x - 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = \dfrac{2}{3}
\end{array} \right.\\
c,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ne 0\\
x \ne 2
\end{array} \right.\\
\dfrac{{x + 1}}{x} - \dfrac{{3x - 1}}{{x - 2}} = 3\\
 \Leftrightarrow \dfrac{{\left( {x + 1} \right)\left( {x - 2} \right) - x\left( {3x - 1} \right)}}{{x\left( {x - 2} \right)}} = 3\\
 \Leftrightarrow \dfrac{{\left( {{x^2} - 2x + x - 2} \right) - \left( {3{x^2} - x} \right)}}{{{x^2} - 2x}} = 3\\
 \Leftrightarrow \dfrac{{{x^2} - x - 2 - 3{x^2} + x}}{{{x^2} - 2x}} = 3\\
 \Leftrightarrow \dfrac{{ - 2{x^2} - 2}}{{{x^2} - 2x}} = 3\\
 \Leftrightarrow  - 2{x^2} - 2 = 3.\left( {{x^2} - 2x} \right)\\
 \Leftrightarrow  - 2{x^2} - 2 = 3{x^2} - 6x\\
 \Leftrightarrow 3{x^2} - 6x + 2{x^2} + 2 = 0\\
 \Leftrightarrow 5{x^2} - 6x + 2 = 0\\
 \Leftrightarrow \left( {5{x^2} - 6x + \dfrac{9}{5}} \right) + \dfrac{1}{5} = 0\\
 \Leftrightarrow 5.\left( {{x^2} - \dfrac{6}{5}x + \dfrac{9}{{25}}} \right) + \dfrac{1}{5} = 0\\
 \Leftrightarrow 5.\left( {{x^2} - 2.x.\dfrac{3}{5} + {{\left( {\dfrac{3}{5}} \right)}^2}} \right) + \dfrac{1}{5} = 0\\
 \Leftrightarrow 5.{\left( {x - \dfrac{3}{5}} \right)^2} + \dfrac{1}{5} = 0\\
5.{\left( {x - \dfrac{3}{5}} \right)^2} \ge 0,\,\,\,\forall x\\
 \Rightarrow 5.{\left( {x - \dfrac{3}{5}} \right)^2} + \dfrac{1}{5} \ge \dfrac{1}{5} > 0,\,\,\,\forall x\\
 \Rightarrow Phương\,\,trình\,\,vô\,\,nghiệm\\
d,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ne  - 1\\
x \ne  - 2
\end{array} \right.\\
\dfrac{2}{{x + 1}} - \dfrac{{6x}}{{x + 2}} + 1 = 0\\
 \Leftrightarrow \dfrac{{2.\left( {x + 2} \right) - 6x\left( {x + 1} \right) + \left( {x + 1} \right)\left( {x + 2} \right)}}{{\left( {x + 1} \right)\left( {x + 2} \right)}} = 0\\
 \Leftrightarrow \dfrac{{2x + 4 - \left( {6{x^2} + 6x} \right) + \left( {{x^2} + 2x + x + 2} \right)}}{{\left( {x + 1} \right)\left( {x + 2} \right)}} = 0\\
 \Leftrightarrow \dfrac{{2x + 4 - 6{x^2} - 6x + {x^2} + 3x + 2}}{{\left( {x + 1} \right)\left( {x + 2} \right)}} = 0\\
 \Leftrightarrow \dfrac{{ - 5{x^2} - x + 6}}{{\left( {x + 1} \right)\left( {x + 2} \right)}} = 0\\
 \Leftrightarrow  - 5{x^2} - x + 6 = 0\\
 \Leftrightarrow 5{x^2} + x - 6 = 0\\
 \Leftrightarrow \left( {5{x^2} - 5x} \right) + \left( {6x - 6} \right) = 0\\
 \Leftrightarrow 5x\left( {x - 1} \right) + 6\left( {x - 1} \right) = 0\\
 \Leftrightarrow \left( {x - 1} \right)\left( {5x + 6} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
5x + 6 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x =  - \dfrac{6}{5}
\end{array} \right.
\end{array}\)