Giải thích các bước giải:

\[\begin{array}{l}
a,\\
2{x^2}\left( {2x - 3} \right) - {x^2}\left( {4{x^2} - 6x + 2} \right) = 0\\
 \Leftrightarrow {x^2}\left( {4x - 6 - \left( {4{x^2} - 6x + 2} \right)} \right) = 0\\
 \Leftrightarrow {x^2}\left( { - 4{x^2} + 10x - 8} \right) = 0\\
 \Leftrightarrow {x^2}\left( { - \left( {2{x^2}} \right) + 2.2x.\frac{5}{2} - \frac{{25}}{4} - \frac{7}{4}} \right) = 0\\
 \Leftrightarrow {x^2}\left( { - {{\left( {2x - \frac{5}{2}} \right)}^2} - \frac{7}{4}} \right) = 0\\
 - {\left( {2x - \frac{5}{2}} \right)^2} - \frac{7}{4} < 0 \Rightarrow x = 0\\
b,\\
{\left( {x - 2} \right)^2} - {\left( {x + 3} \right)^2} - 4\left( {x + 1} \right) = 5\\
 \Leftrightarrow \left( {x - 2 + x + 3} \right)\left( {x - 2 - x - 3} \right) - 4\left( {x + 1} \right) = 5\\
 \Leftrightarrow  - 5\left( {2x + 1} \right) - 4x - 4 = 5\\
 \Leftrightarrow  - 14x = 14\\
 \Leftrightarrow x =  - 1\\
c,\\
4{x^2} - 4x + 1 = {\left( {5 - x} \right)^2}\\
 \Leftrightarrow {\left( {2x - 1} \right)^2} - {\left( {5 - x} \right)^2} = 0\\
 \Leftrightarrow \left( {2x - 1 - 5 + x} \right)\left( {2x - 1 + 5 - x} \right) = 0\\
 \Leftrightarrow \left( {3x - 6} \right)\left( {x + 4} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x =  - 4
\end{array} \right.\\
d,\\
4{x^2} - 8x + 4 = 2\left( {1 - x} \right)\left( {1 + x} \right)\\
 \Leftrightarrow 4{x^2} - 8x + 4 = 2\left( {1 - {x^2}} \right)\\
 \Leftrightarrow 6{x^2} - 8x + 2 = 0\\
 \Leftrightarrow 2\left( {x - 1} \right)\left( {3x - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = \frac{1}{3}
\end{array} \right.
\end{array}\]

\[\begin{array}{l}
a){x^2}(4x - 6 - 4{x^2} + 6x - 2) = 0\\
 \to {x^2}( - 4{x^2} + 10x - 8) = 0\\
 \to {x^2}({x^2} - 5x + 4) = 0\\
 \to {x^2}(x - 1)(x - 4) = 0 \to x = 0;x = 1;x = 4\\
b){x^2} - 4x + 4 - ({x^2} + 6x + 9) - 4x - 4 - 5 = 0\\
 \to  - 14x - 14 = 0 \to x =  - 1
\end{array}\]