tìm gtln.gtnn d=3x-2x^2-4 e=2x^2+5y^2-4xy-4x+2y+3 02/10/2021 Bởi Julia tìm gtln.gtnn d=3x-2x^2-4 e=2x^2+5y^2-4xy-4x+2y+3
\(D=3x-2x^2-4\) \(=-2(x^2-2.x.\dfrac{3}{4}+\dfrac{9}{16})+2.\dfrac{9}{16}-4\) \(=-2(x-\dfrac{3}{4})^2-\dfrac{23}{8}\) Do \((x-\dfrac{3}{4})^2\ge 0\) \(\Rightarrow 2(x-\dfrac{3}{4})^2\ge 0\) \(\Rightarrow -2(x-\dfrac{3}{4})^2\le 0\) \(\Rightarrow -2(x-\dfrac{3}{4})^2-\dfrac{23}{8}\le -\dfrac{23}{8}\) \(\Rightarrow D\le -\dfrac{23}{8}\) Vậy \(D\) đạt giá trị lớn nhất bằng \(-\dfrac{23}{8}\) khi \(x=\dfrac{3}{4}\) Bình luận
\(D=3x-2x^2-4\)
\(=-2(x^2-2.x.\dfrac{3}{4}+\dfrac{9}{16})+2.\dfrac{9}{16}-4\)
\(=-2(x-\dfrac{3}{4})^2-\dfrac{23}{8}\)
Do \((x-\dfrac{3}{4})^2\ge 0\)
\(\Rightarrow 2(x-\dfrac{3}{4})^2\ge 0\)
\(\Rightarrow -2(x-\dfrac{3}{4})^2\le 0\)
\(\Rightarrow -2(x-\dfrac{3}{4})^2-\dfrac{23}{8}\le -\dfrac{23}{8}\)
\(\Rightarrow D\le -\dfrac{23}{8}\)
Vậy \(D\) đạt giá trị lớn nhất bằng \(-\dfrac{23}{8}\) khi \(x=\dfrac{3}{4}\)