Đáp án:
Min=1
Giải thích các bước giải:
\(\begin{array}{l}
\dfrac{{{x^2} - 7x + 15}}{{x - 1}} = \dfrac{{{x^2} - 2x + 1 - 5x + 14}}{{x - 1}}\\
= \dfrac{{{{\left( {x - 1} \right)}^2} - 5\left( {x - 1} \right) + 9}}{{x - 1}}\\
= \left( {x - 1} \right) - 5 + \dfrac{9}{{x - 1}}\\
= \left( {x - 1} \right) + \dfrac{9}{{x - 1}} - 5\\
Do:x > 1\\
BDT:Co - si:\left( {x - 1} \right) + \dfrac{9}{{x - 1}} \ge 2\sqrt {\left( {x - 1} \right).\dfrac{9}{{x - 1}}} \\
\to \left( {x - 1} \right) + \dfrac{9}{{x - 1}} \ge 6\\
\to \left( {x - 1} \right) + \dfrac{9}{{x - 1}} - 5 \ge 1\\
\to Min = 1\\
\Leftrightarrow \left( {x - 1} \right) = \dfrac{9}{{x - 1}}\\
\Leftrightarrow {\left( {x - 1} \right)^2} = 9\\
\to \left[ \begin{array}{l}
x - 1 = 3\\
x - 1 = - 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 4\\
x = - 2\left( l \right)
\end{array} \right.
\end{array}\)
