\(A=8x-x^2=16-\left(x^2-8x+16\right)=16-\left(x-4\right)^2\le16;\forall x\)
\(\Rightarrow A_{max}=16\Leftrightarrow x-4=0\Leftrightarrow x=4\)
\(B=12x-4x^2-5\)\(=4-\left(4x^2-2.3.2x+9\right)=4-\left(2x-3\right)^2\le4;\forall x\)
\(\Rightarrow B_{max}=4\Leftrightarrow2x-3=0\Leftrightarrow x=\dfrac{3}{2}\)
\(C=\dfrac{2x^2+9}{x^2+4}=\dfrac{2\left(x^2+4\right)}{x^2+4}+\dfrac{1}{x^2+4}=2+\dfrac{1}{x^2+4}\)
Có \(x^2+4\ge4;\forall x\) \(\Rightarrow\dfrac{1}{x^2+4}\le\dfrac{1}{4}\) \(\Rightarrow C\le\dfrac{1}{4}+2=\dfrac{9}{4}\)
\(\Rightarrow C_{max}=\dfrac{9}{4}\Leftrightarrow x^2=0\Leftrightarrow x=0\)
Đúng 1
Bình luận (1)
