Câu hỏi của Nguyễn Ngọc Ánh

Question

Tìm GTLN của:

$A=13\sqrt{x^2-x^4}+9\sqrt{x^2+x^4}$ với $0\le x\le1$

Nguyễn Việt Lâm · Accepted Answer

$A^2=\left(\sqrt{13}.\sqrt{13x^2-13x^4}+3\sqrt{3}.\sqrt{3x^2+3x^4}\right)^2$

$\Rightarrow A^2\le\left(13+27\right)\left(16x^2-10x^4\right)=40\left[\frac{32}{5}-10\left(x^2-\frac{4}{5}\right)^2\right]\le256$

$\Rightarrow A\le16\Rightarrow A_{max}=16$ khi $x^2=\frac{4}{5}$

Câu trả lời:

$A^2=\left(\sqrt{13}.\sqrt{13x^2-13x^4}+3\sqrt{3}.\sqrt{3x^2+3x^4}\right)^2$

$\Rightarrow A^2\le\left(13+27\right)\left(16x^2-10x^4\right)=40\left[\frac{32}{5}-10\left(x^2-\frac{4}{5}\right)^2\right]\le256$

$\Rightarrow A\le16\Rightarrow A_{max}=16$ khi $x^2=\frac{4}{5}$