Đáp án:
c. \(\left[ \begin{array}{l}
x = \dfrac{1}{5}\\
x = - 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{*{20}{l}}
{a.\left( {x - 1} \right)\left( {x - 2} \right) - \left( {x - 1} \right)\left( {2x + 4} \right) = - \left( {x - 1} \right)\left( {x + 1} \right)}\\
{ \to \left( {x - 1} \right)\left( {x - 2} \right) - \left( {x - 1} \right)\left( {2x + 4} \right) + \left( {x - 1} \right)\left( {x + 1} \right) = 0}\\
{ \to \left( {x - 1} \right)\left( {x - 2 - 2x - 4 + x + 1} \right) = 0}\\
{ \to - 5\left( {x - 1} \right) = 0}\\
{ \to x = 1}\\
{b.6 - 4x - 9x + 6{x^2} - 6{x^2} + 30x = 7}\\
{ \to 17x = 1}\\
{ \to x = \dfrac{1}{{17}}}\\
{c.3 - 5x + 6x - 10{x^2} - 4{x^2} - 12x = {x^2} + x}\\
{ \to 15{x^2} + 12x - 3 = 0}\\
{ \to \left[ {\begin{array}{*{20}{l}}
{x = \dfrac{1}{5}}\\
{x = - 1}
\end{array}} \right.}
\end{array}\)
