Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
{\left( {5{x^2} - 2x + 10} \right)^2} = {\left( {3{x^2} + 10x - 8} \right)^2}\\
\Leftrightarrow \left[ \begin{array}{l}
5{x^2} - 2x + 10 = 3{x^2} + 10x - 8\\
5{x^2} - 2x + 10 = - 3{x^2} - 10x + 8
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2{x^2} - 12x + 18 = 0\\
8{x^2} + 8x + 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = \frac{{ - 1}}{2}
\end{array} \right.\\
b,\\
16{x^2} - 8x + 1 = 4\left( {x + 3} \right)\left( {4x - 1} \right)\\
\Leftrightarrow {\left( {4x - 1} \right)^2} = 4\left( {x + 3} \right)\left( {4x - 1} \right)\\
\Leftrightarrow \left( {4x - 1} \right).\left[ {\left( {4x - 1} \right) - 4\left( {x + 3} \right)} \right] = 0\\
\Leftrightarrow \left( {4x - 1} \right).\left( { - 13} \right) = 0\\
\Leftrightarrow 4x - 1 = 0\\
\Leftrightarrow x = \frac{1}{4}\\
c,\\
27{x^2}\left( {x - 3} \right) - 12\left( {{x^2} + 3x} \right) = 0\\
\Leftrightarrow 9{x^2}\left( {x - 3} \right) - 4x\left( {x + 3} \right) = 0\\
\Leftrightarrow x.\left[ {9x\left( {x - 3} \right) - 4\left( {x + 3} \right)} \right] = 0\\
\Leftrightarrow x.\left[ {9{x^2} - 27x - 4x - 12} \right] = 0\\
\Leftrightarrow x.\left( {9{x^2} - 31x - 12} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = ...
\end{array} \right.\\
d,\\
2\left( {9{x^2} + 6x + 1} \right) = \left( {3x + 1} \right)\left( {x - 2} \right)\\
\Leftrightarrow 2{\left( {3x + 1} \right)^2} - \left( {3x + 1} \right)\left( {x - 2} \right) = 0\\
\Leftrightarrow \left( {3x + 1} \right).\left[ {2\left( {3x + 1} \right) - \left( {x - 2} \right)} \right] = 0\\
\Leftrightarrow \left( {3x + 1} \right).\left( {5x + 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = - \frac{1}{3}\\
x = - \frac{4}{5}
\end{array} \right.
\end{array}\)
