Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\left( {x - 3} \right)\left( {x - 5} \right) < 0\\
TH1:\,\,x \le 3 \Rightarrow \left\{ \begin{array}{l}
x - 3 \le 0\\
x - 5 < 0
\end{array} \right. \Rightarrow \left( {x - 3} \right)\left( {x - 5} \right) \ge 0\\
TH2:\,\,\,x \ge 5 \Rightarrow \left\{ \begin{array}{l}
x - 3 > 0\\
x - 5 \ge 0
\end{array} \right. \Rightarrow \left( {x - 3} \right)\left( {x - 5} \right) \ge 0\\
TH3:\,\,3 < x < 5 \Rightarrow \left\{ \begin{array}{l}
x - 3 > 0\\
x - 5 < 0
\end{array} \right. \Rightarrow \left( {x - 3} \right)\left( {x - 5} \right) < 0\\
x \in Z \Rightarrow x = 4\\
b,\\
xy - 2x + 3y = 1\\
\Leftrightarrow x\left( {y - 2} \right) + 3\left( {y - 2} \right) = 1 - 6\\
\Leftrightarrow \left( {x + 3} \right)\left( {y - 2} \right) = - 5\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 3 = 1\\
y - 2 = - 5
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 3 = - 1\\
y - 1 = 5
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 3 = - 5\\
y - 1 = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
y + 3 = 5\\
y - 1 = - 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = - 2\\
y = - 3
\end{array} \right.\\
\left\{ \begin{array}{l}
x = - 4\\
y = 6
\end{array} \right.\\
\left\{ \begin{array}{l}
x = - 8\\
y = 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 2\\
y = 0
\end{array} \right.
\end{array} \right.\\
c,\\
{2^{x + 3}} + {2^{x + 4}} = 192\\
\Leftrightarrow {2^{x + 3}}\left( {1 + 2} \right) = 192\\
\Leftrightarrow {3.2^{x + 3}} = 192\\
\Leftrightarrow {2^{x + 3}} = 64\\
\Leftrightarrow x + 3 = 6\\
\Leftrightarrow x = 3\\
d,\\
{2^x}{.3^x} = {4.3^2}\\
\Leftrightarrow {6^x} = {6^2}\\
\Leftrightarrow x = 2
\end{array}\)
