Toán Tìm gtnn của biểu thức sau :2(x-2)⁴ -3x² +12x -1 14/09/2021 By Quinn Tìm gtnn của biểu thức sau :2(x-2)⁴ -3x² +12x -1
Đáp án: $\begin{array}{l}A = 2{\left( {x – 2} \right)^4} – 3{x^2} + 12x – 1\\ = 2{\left( {x – 2} \right)^4} – 3.\left( {{x^2} – 4x + 4} \right) + 12 – 1\\ = 2{\left( {x – 2} \right)^4} – 3{\left( {x – 2} \right)^2} + 11\\Đặt:{\left( {x – 2} \right)^2} = t\left( {t \ge 0} \right)\\ \Rightarrow A = 2.{t^2} – 3t + 11\\ = 2.\left( {{t^2} – 2.t.\dfrac{3}{4} + \dfrac{9}{{16}}} \right) – 2.\dfrac{9}{{16}} + 11\\ = 2.{\left( {t – \dfrac{3}{4}} \right)^2} + \dfrac{{79}}{8} \ge \dfrac{{79}}{8}\\ \Rightarrow GTNN:A = \dfrac{{79}}{8} \Leftrightarrow t = \dfrac{3}{4}\\ \Rightarrow {\left( {x – 2} \right)^2} = \dfrac{3}{4}\\ \Rightarrow x = 2 \pm \dfrac{{\sqrt 3 }}{2}\end{array}$ Trả lời
Đáp án:
$\begin{array}{l}
A = 2{\left( {x – 2} \right)^4} – 3{x^2} + 12x – 1\\
= 2{\left( {x – 2} \right)^4} – 3.\left( {{x^2} – 4x + 4} \right) + 12 – 1\\
= 2{\left( {x – 2} \right)^4} – 3{\left( {x – 2} \right)^2} + 11\\
Đặt:{\left( {x – 2} \right)^2} = t\left( {t \ge 0} \right)\\
\Rightarrow A = 2.{t^2} – 3t + 11\\
= 2.\left( {{t^2} – 2.t.\dfrac{3}{4} + \dfrac{9}{{16}}} \right) – 2.\dfrac{9}{{16}} + 11\\
= 2.{\left( {t – \dfrac{3}{4}} \right)^2} + \dfrac{{79}}{8} \ge \dfrac{{79}}{8}\\
\Rightarrow GTNN:A = \dfrac{{79}}{8} \Leftrightarrow t = \dfrac{3}{4}\\
\Rightarrow {\left( {x – 2} \right)^2} = \dfrac{3}{4}\\
\Rightarrow x = 2 \pm \dfrac{{\sqrt 3 }}{2}
\end{array}$