Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
x.\left( {6x - 5} \right) - 6{x^2} + 3x - 4 = 0\\
\Leftrightarrow \left( {6{x^2} - 5x} \right) - 6{x^2} + 3x - 4 = 0\\
\Leftrightarrow 6{x^2} - 5x - 6{x^2} + 3x - 4 = 0\\
\Leftrightarrow - 2x - 4 = 0\\
\Leftrightarrow 2x = - 4\\
\Leftrightarrow x = - 2\\
b,\\
{\left( {x - 2} \right)^2} - \left( {2x - 1} \right)\left( {x + 3} \right) = - {x^2} + 4\\
\Leftrightarrow \left( {{x^2} - 4x + 4} \right) - \left( {2{x^2} + 5x - 3} \right) + {x^2} - 4 = 0\\
\Leftrightarrow {x^2} - 4x + 4 - 2{x^2} - 5x + 3 + {x^2} - 4 = 0\\
\Leftrightarrow - 9x + 3 = 0\\
\Leftrightarrow 9x = 3\\
\Leftrightarrow x = \dfrac{1}{3}\\
c,\\
\left( {3x + 2} \right)\left( {x - 3} \right) + 9 = {x^2}\\
\Leftrightarrow \left( {3x + 2} \right)\left( {x - 3} \right) - {x^2} + 9 = 0\\
\Leftrightarrow \left( {3x + 2} \right)\left( {x - 3} \right) - \left( {{x^2} - 9} \right) = 0\\
\Leftrightarrow \left( {3x + 2} \right)\left( {x - 3} \right) - \left( {x - 3} \right)\left( {x + 3} \right) = 0\\
\Leftrightarrow \left( {x - 3} \right).\left[ {\left( {3x + 2} \right) - \left( {x + 3} \right)} \right] = 0\\
\Leftrightarrow \left( {x - 3} \right)\left( {2x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 3 = 0\\
2x - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = \dfrac{1}{2}
\end{array} \right.
\end{array}\)
