Toán Lớp 8: Tìm giá trị nhỏ nhất của biểu thức:
(x-2) (x-3) (x-4) (x-5)
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Toán Lớp 8: Tìm giá trị nhỏ nhất của biểu thức:
(x-2) (x-3) (x-4) (x-5)
Comments ( 1 )
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}\)