Đáp án:
\[{A_{\min }} = - 1 \Leftrightarrow x = \dfrac{{7 \pm \sqrt 5 }}{2}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = \left( {x - 2} \right)\left( {x - 3} \right)\left( {x - 4} \right)\left( {x - 5} \right)\\
= \left[ {\left( {x - 2} \right)\left( {x - 5} \right)} \right].\left[ {\left( {x - 3} \right)\left( {x - 4} \right)} \right]\\
= \left( {{x^2} - 5x - 2x + 10} \right).\left( {{x^2} - 4x - 3x + 12} \right)\\
= \left( {{x^2} - 7x + 10} \right).\left( {{x^2} - 7x + 12} \right)\\
= \left[ {\left( {{x^2} - 7x + 11} \right) - 1} \right].\left[ {\left( {{x^2} - 7x + 11} \right) + 1} \right]\\
= {\left( {{x^2} - 7x + 11} \right)^2} - {1^2}\\
= {\left( {{x^2} - 7x + 11} \right)^2} - 1\\
{\left( {{x^2} - 7x + 11} \right)^2} \ge 0,\,\,\,\forall x\\
\Rightarrow {\left( {{x^2} - 7x + 11} \right)^2} - 1 \ge - 1,\,\,\,\,\forall x\\
\Rightarrow A \ge - 1,\,\,\,\forall x\\
\Rightarrow {A_{\min }} = - 1 \Leftrightarrow {\left( {{x^2} - 7x + 11} \right)^2} = 0 \Leftrightarrow {x^2} - 7x + 11 = 0\\
\Leftrightarrow \left( {{x^2} - 7x + \dfrac{{49}}{4}} \right) - \dfrac{5}{4} = 0\\
\Leftrightarrow \left( {{x^2} - 2.x.\dfrac{7}{2} + {{\left( {\dfrac{7}{2}} \right)}^2}} \right) = \dfrac{5}{4}\\
\Leftrightarrow {\left( {x - \dfrac{7}{2}} \right)^2} = \dfrac{5}{4}\\
\Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{7}{2} = \dfrac{{\sqrt 5 }}{2}\\
x - \dfrac{7}{2} = - \dfrac{{\sqrt 5 }}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{7 + \sqrt 5 }}{2}\\
x = \dfrac{{7 - \sqrt 5 }}{2}
\end{array} \right.\\
Vậy\,\,\,{A_{\min }} = - 1 \Leftrightarrow x = \dfrac{{7 \pm \sqrt 5 }}{2}
\end{array}\)
