Đáp án:
h. \(\left[ \begin{array}{l}
x = 3\\
x = - 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
d.6 - 9\left( {x + 1} \right) - 4\left( {x - 1} \right) = 3 - 6x\\
\to 6 - 9x - 9 - 4x + 4 = 3 - 6x\\
\to 7x = - 2\\
\to x = - \frac{2}{7}\\
e.\left( {x + 5} \right)\left( {x + 5 + 3x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x + 5 = 0\\
4x + 4 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 5\\
x = - 1
\end{array} \right.\\
f.\left( {9 - x} \right)\left( {3 + 2x - 4 + x} \right) = 0\\
\to \left[ \begin{array}{l}
9 - x = 0\\
3x - 1 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 9\\
x = \frac{1}{3}
\end{array} \right.\\
h.DK:x \ne \pm 2\\
Pt \to \left( {x - 1} \right)\left( {x - 2} \right) + 2\left( {x + 2} \right) = 12\\
\to {x^2} - 3x + 2 + 2x + 4 = 12\\
\to {x^2} - x - 6 = 0\\
\to \left( {x - 3} \right)\left( {x + 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = 3\\
x = - 2
\end{array} \right.
\end{array}\)
