Đáp án:

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

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
3,\\
a,\\
\left( {2 - 3x} \right)\left( {{x^2} + 1} \right) + 3x = 2\\
 \Leftrightarrow \left( {2 - 3x} \right)\left( {{x^2} + 1} \right) + 3x - 2 = 0\\
 \Leftrightarrow \left( {2 - 3x} \right)\left( {{x^2} + 1} \right) - \left( {2 - 3x} \right) = 0\\
 \Leftrightarrow \left( {2 - 3x} \right).\left[ {\left( {{x^2} + 1} \right) - 1} \right] = 0\\
 \Leftrightarrow \left( {2 - 3x} \right).{x^2} = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
2 - 3x = 0\\
{x^2} = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{2}{3}\\
x = 0
\end{array} \right.\\
b,\\
{x^2} - 1 + \left( {x - 1} \right)\left( {x + 2} \right) = 0\\
 \Leftrightarrow {x^2} - {1^2} + \left( {x - 1} \right)\left( {x + 2} \right) = 0\\
 \Leftrightarrow \left( {x - 1} \right)\left( {x + 1} \right) + \left( {x - 1} \right)\left( {x + 2} \right) = 0\\
 \Leftrightarrow \left( {x - 1} \right).\left[ {\left( {x + 1} \right) + \left( {x + 2} \right)} \right] = 0\\
 \Leftrightarrow \left( {x - 1} \right)\left( {2x + 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
2x + 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x =  - \dfrac{3}{2}
\end{array} \right.\\
c,\\
{x^3} + 3{x^2} + 3x + 1 = 0\\
 \Leftrightarrow {x^3} + 3.{x^2}.1 + 3.x{.1^2} + {1^3} = 0\\
 \Leftrightarrow {\left( {x + 1} \right)^3} = 0\\
 \Leftrightarrow x + 1 = 0\\
 \Leftrightarrow x =  - 1\\
d,\\
27{x^3} = 1\\
 \Leftrightarrow {x^3} = \dfrac{1}{{27}}\\
 \Leftrightarrow {x^3} = {\left( {\dfrac{1}{3}} \right)^3}\\
 \Leftrightarrow x = \dfrac{1}{3}\\
4,\\
a,\\
{\left( {x - 1} \right)^2} = {\left( {2 - 4x} \right)^2}\\
 \Leftrightarrow {\left( {x - 1} \right)^2} - {\left( {2 - 4x} \right)^2} = 0\\
 \Leftrightarrow \left[ {\left( {x - 1} \right) - \left( {2 - 4x} \right)} \right].\left[ {\left( {x - 1} \right) + \left( {2 - 4x} \right)} \right] = 0\\
 \Leftrightarrow \left( {x - 1 - 2 + 4x} \right).\left( {x - 1 + 2 - 4x} \right) = 0\\
 \Leftrightarrow \left( {5x - 3} \right).\left( {1 - 3x} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
5x - 3 = 0\\
1 - 3x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
5x = 3\\
3x = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{3}{5}\\
x = \dfrac{1}{3}
\end{array} \right.\\
b,\\
{\left( {5x - 1} \right)^3} - 8{x^3} = 0\\
 \Leftrightarrow {\left( {5x - 1} \right)^3} = 8{x^3}\\
 \Leftrightarrow {\left( {5x - 1} \right)^3} = {\left( {2x} \right)^3}\\
 \Leftrightarrow 5x - 1 = 2x\\
 \Leftrightarrow 3x - 1 = 0\\
 \Leftrightarrow 3x = 1\\
 \Leftrightarrow x = \dfrac{1}{3}
\end{array}\)