Giải thích các bước giải:
\(a.A = \frac{{x - 3 + 10}}{{x - 3}} = 1 + \frac{{10}}{{x - 3}}\)
Để A min
⇔ \(\frac{{10}}{{x - 3}}\) nguyên
⇔\(\left( {x - 3} \right) \in U\left( {10} \right)\)
Xét:
\(\left[ \begin{array}{l}
x - 3 = 10\\
x - 3 = - 10\\
x - 3 = 5\\
x - 3 = - 5\\
x - 3 = 2\\
x - 3 = - 2\\
x - 3 = 1\\
x - 3 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 13\\
x = - 7\\
x = 8\\
x = - 2\\
x = 5\\
x = 1\\
x = 4\\
x = 2
\end{array} \right. \to \left[ \begin{array}{l}
A = 2\\
A = 0\\
A = 3\\
A = - 1\\
A = 6\\
A = - 4\\
A = 11\\
A = - 9
\end{array} \right.\)
\( \to {A_{\min }} = - 9 \Leftrightarrow x = 2\)
\(b.B = \frac{{x - 1 + 5}}{{x - 1}} = 1 + \frac{5}{{x - 1}}\)
Để B min
⇔\(\frac{5}{{x - 1}}\) min
\(\begin{array}{l}
\Leftrightarrow \left( {x - 1} \right) \in U\left( 5 \right);{\left( {x - 1} \right)_{\min }}\\
\to x - 1 = - 1 \to x = 0\\
\to A = - 4
\end{array}\)
