Đáp án:
36) \(\left[ \begin{array}{l}
x = - 16\\
x = 14
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
20) - 12\left( {x - 5} \right) + 7\left( {3 - x} \right){\rm{ = }}5\\
\to - 12x + 60 + 21 - 7x = 5\\
\to 19x = - 76\\
\to x = - 4\\
21)\left( {x - 2} \right).\left( {x + 4} \right) = 0\\
\to \left[ \begin{array}{l}
x - 2 = 0\\
x + 4 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
x = - 4
\end{array} \right.\\
22)\left( {x - 2} \right)\left( {x + 15} \right) = 0\\
\to \left[ \begin{array}{l}
x - 2 = 0\\
x + 15 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
x = - 15
\end{array} \right.\\
23)\left( {7 - x} \right)\left( {x + 19} \right) = 0\\
\to \left[ \begin{array}{l}
7 - x = 0\\
x + 19 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 7\\
x = - 19
\end{array} \right.\\
24) - 5 < x < 1\\
\to x \in \left\{ { - 4; - 3; - 2; - 1;0} \right\}\\
25)\left| x \right| < 3\\
\to - 3 < x < 3\\
\to x \in \left\{ { - 2; - 1;0;1;2} \right\}\\
26)\left( {x - 3} \right)\left( {x - 5} \right) = 0\\
\to \left[ \begin{array}{l}
x - 3 = 0\\
x - 5 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = 5
\end{array} \right.\\
27)\left( {x - 5} \right)\left( {x - 7} \right) = 0\\
\to \left[ \begin{array}{l}
x - 5 = 0\\
x - 7 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 5\\
x = 7
\end{array} \right.\\
29) - 6x + 7 = 25\\
\to - 6x = 18\\
\to x = - 3\\
30)46 - x + 11 = - 48\\
\to x = 105\\
31)\left| {x - 12} \right| = 20\\
\to \left[ \begin{array}{l}
x - 12 = 20\\
x - 12 = - 20
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 32\\
x = - 8
\end{array} \right.\\
34)\left( {x - 11} \right)\left( {x + 5} \right) = 0\\
\to \left[ \begin{array}{l}
x = 11\\
x = - 5
\end{array} \right.\\
36)\left| {x + 1} \right| = 15\\
\to \left[ \begin{array}{l}
x + 1 = - 15\\
x + 1 = 15
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 16\\
x = 14
\end{array} \right.
\end{array}\)
