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Ta có: \(2x^2+\frac{y^2}{4}+\frac{1}{x^2}=4\)
=> \(\left(x^2+\frac{y^2}{4}\right)+\left(x^2+\frac{1}{x^2}\right)=4\)
Lại có: \(x^2+\frac{y^2}{4}\ge2.x.\frac{y}{2}=xy\) Và \(x^2+\frac{1}{x^2}\ge2.x.\frac{1}{x}=2\)
=> \(4\ge xy+2\)=> \(2\ge xy\)
=> \(A=2016+xy\le2016+2=2018\)
=> Amin=2018
\(\sqrt[]{\sqrt{ }\frac{ }{ }\sqrt[]{}3\hept{\begin{cases}\\\\\end{cases}}3\frac{ }{ }\sqrt{ }\cos\hept{\begin{cases}\\\\\end{cases}}\Omega3\cong}\)
Ta có :
\(2x^2+\dfrac{1}{x^2}+\dfrac{y^2}{4}=4\)
\(\Rightarrow\left(x^2+\dfrac{1}{x^2}-2\right)+\left(x^2+\dfrac{y^2}{4}-xy\right)=2-xy\)
\(\Rightarrow\left(x-\dfrac{1}{x}\right)^2+\left(x-\dfrac{y}{2}\right)^2=2-xy\)
Ta có:
\(\left(x-\dfrac{1}{x}\right)^2\ge0\forall x\)
\(\left(x-\dfrac{y}{2}\right)^2\ge0\forall x,y\)
\(\Rightarrow\left(x-\dfrac{1}{x}\right)^2+\left(x-\dfrac{y}{2}\right)^2\ge0\forall x,y\)
\(\Rightarrow2-xy\ge0\forall x,y\)
\(\Rightarrow xy\le2\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}x-\dfrac{1}{x}=0\\x-\dfrac{y}{2}=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{x}\\x=\dfrac{y}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x^2=1\\y=2x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\\\left[{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\end{matrix}\right.\)
Vậy (x;y) nguyên thỏa mãn là : (1;2);(-1;-2)
Đề bài sai, đề đúng thì phân thức đằng sau dấu chia phải là:
\(\dfrac{4x^4+4x^2y+y^2-4}{x^2+y+xy+x}\)
Câu 1:
Áp dụng BĐT Cô-si:
\(x^4+y^2\geq 2\sqrt{x^4y^2}=2x^2y\Rightarrow \frac{x}{x^4+y^2}\leq \frac{x}{2x^2y}=\frac{1}{2xy}=\frac{1}{2}(1)\)
\(x^2+y^4\geq 2\sqrt{x^2y^4}=2xy^2\Rightarrow \frac{y}{x^2+y^4}\leq \frac{y}{2xy^2}=\frac{1}{2xy}=\frac{1}{2}(2)\)
Lấy \((1)+(2)\Rightarrow A\leq \frac{1}{2}+\frac{1}{2}=1\)
Vậy \(A_{\max}=1\). Dấu bằng xảy ra khi \(x=y=1\)
Câu 2:
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)(x^2+y^2+2xy)\geq (1+1)^2\)
\(\Rightarrow \frac{1}{x^2+y^2}+\frac{1}{2xy}\geq \frac{4}{x^2+y^2+2xy}=\frac{4}{(x+y)^2}\geq \frac{4}{1}=4(*)\)
(do \(x+y\leq 1\) )
Áp dụng BĐT Cô-si:
\(\frac{1}{4xy}+4xy\geq 2\sqrt{\frac{4xy}{4xy}}=2(**)\)
\(x+y\geq 2\sqrt{xy}\Leftrightarrow 1\geq 2\sqrt{xy}\Rightarrow xy\leq \frac{1}{4}\)
\(\Rightarrow \frac{5}{4xy}\geq \frac{5}{4.\frac{1}{4}}=5(***)\)
Cộng \((*)+(**)+(***)\Rightarrow B\geq 4+2+5=11\)
Vậy \(B_{\min}=11\)
Dấu bằng xảy ra khi \(x=y=\frac{1}{2}\)
\(2x^2+\frac{1}{x^2}+\frac{y^2}{4}=4\)
\(\Leftrightarrow\left(x^2+\frac{1}{x^2}-2\right)+\left(x^2+\frac{y^2}{4}+xy\right)=2+xy\)\(\Leftrightarrow\left(x-\frac{1}{x}\right)^2+\left(x+\frac{y}{2}\right)^2=2+xy\)
\(\Rightarrow2+xy\ge0\)\(\Rightarrow xy\ge-2\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-\frac{1}{x}\right)^2=0\\\left(x+\frac{y}{2}\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\end{matrix}\right.\)
Ta có:
\(2x^2+\dfrac{1}{x^2}+\dfrac{y^2}{4}=4\)
\(\Leftrightarrow2=x^2-2+\dfrac{1}{x^2}+x^2-xy+\dfrac{y^2}{4}+xy\)
\(\Leftrightarrow2=\left(x-\dfrac{1}{x}\right)^2+\left(x-\dfrac{y}{2}\right)^2+xy\)
Vì : \(\left(x-\dfrac{1}{x}\right)^2+\left(x-\dfrac{y}{2}\right)^2\ge0\)
\(\Rightarrow xy\le2\)
Vậy GTLN của xy=2 \(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{x}=0\\x-\dfrac{y}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\Rightarrow y=2\\x=-1\Rightarrow y=-2\end{matrix}\right.\)