Giải thích các bước giải:

\(\begin{array}{l}
a,\\
{\left( {x - 1} \right)^2} + {\left( {x + 1} \right)^2} + \left( {x - 1} \right)\left( {x + 1} \right) = 3{x^2}\\
 \Leftrightarrow \left( {{x^2} - 2x + 1} \right) + \left( {{x^2} + 2x + 1} \right) + \left( {{x^2} - {1^2}} \right) = 3{x^2}\\
 \Leftrightarrow {x^2} - 2x + 1 + {x^2} + 2x + 1 + {x^2} - 1 = 3{x^2}\\
 \Leftrightarrow 3{x^2} + 1 = 3{x^2}\\
 \Leftrightarrow 1 = 0\,\,\,\,\,\,\left( {vn} \right)\\
b,\\
{\left( {x - 2} \right)^2} - {\left( {x - 3} \right)^2} = 5\\
 \Leftrightarrow \left[ {\left( {x - 2} \right) - \left( {x - 3} \right)} \right].\left[ {\left( {x - 2} \right) + \left( {x - 3} \right)} \right] = 5\\
 \Leftrightarrow \left( {x - 2 - x + 3} \right).\left( {x - 2 + x - 3} \right) = 5\\
 \Leftrightarrow 1.\left( {2x - 5} \right) = 5\\
 \Leftrightarrow 2x - 5 = 5\\
 \Leftrightarrow x = 5\\
c,\\
{\left( {x + 1} \right)^2}{\left( {x + 3} \right)^2} = {27^2}\\
 \Leftrightarrow {\left[ {\left( {x + 1} \right)\left( {x + 3} \right)} \right]^2} = {27^2}\\
 \Leftrightarrow {\left( {{x^2} + 4x + 3} \right)^2} = {27^2}\\
 \Leftrightarrow \left[ \begin{array}{l}
{x^2} + 4x + 3 = 27\\
{x^2} + 4x + 3 =  - 27
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
{x^2} + 4x + 3 + 1 = 27 + 1\\
{x^2} + 4x + 3 + 1 =  - 27 + 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
{x^2} + 4x + 4 = 28\\
{x^2} + 4x + 4 =  - 26
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
{\left( {x + 2} \right)^2} = 28\\
{\left( {x + 2} \right)^2} =  - 26\,\,\,\,\,\left( L \right)
\end{array} \right.\\
 \Leftrightarrow {\left( {x + 2} \right)^2} = 28 \Leftrightarrow \left[ \begin{array}{l}
x + 2 = \sqrt {28} \\
x + 2 =  - \sqrt {28} 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\sqrt 7  - 2\\
x =  - 2\sqrt 7  - 2
\end{array} \right.\\
d,\\
{\left( {2x - 1} \right)^2} + {\left( {2x + 5} \right)^2} - 4{x^2} = 7\\
 \Leftrightarrow \left( {4{x^2} - 4x + 1} \right) + \left( {4{x^2} + 20x + 25} \right) - 4{x^2} = 7\\
 \Leftrightarrow 4{x^2} + 16x + 26 = 7\\
 \Leftrightarrow 4{x^2} + 16x + 19 = 0\\
 \Leftrightarrow \left( {4{x^2} + 16x + 16} \right) + 3 = 0\\
 \Leftrightarrow 4.\left( {{x^2} + 4x + 4} \right) + 3 = 0\\
 \Leftrightarrow 4.{\left( {x + 2} \right)^2} =  - 3\,\,\,\,\,\left( {vn} \right)\\
e,\\
\left( {2x - 3} \right)\left( {2x + 3} \right) - 4{x^2} = 2x\\
 \Leftrightarrow \left[ {{{\left( {2x} \right)}^2} - {3^2}} \right] - 4{x^2} = 2x\\
 \Leftrightarrow 4{x^2} - 9 - 4{x^2} = 2x\\
 \Leftrightarrow  - 9 = 2x\\
 \Leftrightarrow x =  - \dfrac{9}{2}
\end{array}\)