Ta có: \(8-y^2=\left|xy-4\right|\ge0\Rightarrow y^2\le8\) (1)

\(x^2+2=xy\Rightarrow x^2-xy+2=0\)

\(\Leftrightarrow\left(x-\dfrac{y}{2}\right)^2-\dfrac{y^2}{4}+2=0\Leftrightarrow\dfrac{y^2}{4}-2=\left(x-\dfrac{y}{2}\right)^2\ge0\)

\(\Rightarrow y^2\ge8\) (2)

Từ (1); (2) \(\Rightarrow y^2=8\)

Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}y^2=8\\xy-4=0\\x-\dfrac{y}{2}=0\end{matrix}\right.\) \(\Leftrightarrow...\)

\(\hept{\begin{cases}x^2+y^2+xy=3\\xy+3x^2=4\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}4\left(x^2+y^2+xy\right)=3\left(3x^2+xy\right)\text{ }\left(\text{1}\right)\\3x^2+xy=4\end{cases}}\)

\(\left(1\right)\Leftrightarrow5x^2-xy-4y^2=0\Leftrightarrow\left(x-y\right)\left(5x+4y\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x-y=0\\5x+4y=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}y=x\\y=-\frac{5}{4}x\end{cases}}\)

\(\text{TH1:}y=x\), ta được hệ \(\hept{\begin{cases}x=y\\3x^2+xy=4\end{cases}}\)

TH2: \(y=-\frac{5}{4}x\), ta có hệ \(\hept{\begin{cases}y=-\frac{5}{4}x\\3x^2+xy=4\end{cases}}\)

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\(\left\{{}\begin{matrix}\left(x-1\right)\left(y+1\right)=xy-1\\\left(x-2\right)\left(y-2\right)=xy-8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}xy+x-y-1=xy-1\\xy-2x-2y+4=xy-8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x-y=0\\-2x-2y=-12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-2y=0\\2x+2y=12\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4x=12\\x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=x=3\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\left(x-1\right)\left(y+1\right)=xy-1\\\left(x-2\right)\left(y-2\right)=xy-8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy+x-y-1=xy-1\\xy-2x-2y+4=xy-8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\x+y=6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+x+y^2+y=8\\\left(x^2+x\right)\left(y^2+y\right)=12\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x^2+x=u\\y^2+y=v\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u+v=8\\uv=12\end{matrix}\right.\) \(\Rightarrow\left(u;v\right)=\left(6;2\right);\left(2;6\right)\)

TH1: \(\left\{{}\begin{matrix}x^2+x=6\\y^2+y=2\end{matrix}\right.\) \(\Rightarrow...\)

TH2: ... tương tự

đkxđ: \(\left\{{}\begin{matrix}x\ne0\\y\ne0\end{matrix}\right.\)

pt đầu \(\Leftrightarrow x+\dfrac{2}{x}+y+\dfrac{1}{y}=6\)            (3)

pt thứ 2 \(\Leftrightarrow x^2+\dfrac{4}{x^2}+y^2+\dfrac{1}{y^2}=14\) \(\Leftrightarrow\left(x^2+2.x.\dfrac{2}{x}+\dfrac{4}{x^2}\right)+\left(y^2+2y.\dfrac{1}{y}+\dfrac{1}{y^2}\right)=20\)

\(\Leftrightarrow\left(x+\dfrac{2}{x}\right)^2+\left(y+\dfrac{1}{y}\right)^2=20\)                   (4)

Đặt \(\left\{{}\begin{matrix}x+\dfrac{2}{x}=u\left(\left|u\right|\ge2\sqrt{2}\right)\\y+\dfrac{1}{y}=v\left(\left|v\right|\ge2\right)\end{matrix}\right.\) thì từ (3) và (4) suy ra \(\left\{{}\begin{matrix}u+v=6\\u^2+v^2=20\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}v=6-u\\u^2+\left(6-u\right)^2=20\end{matrix}\right.\) 

\(u^2+\left(6-u\right)^2=20\) \(\Leftrightarrow u^2+36-12u+u^2=20\) \(\Leftrightarrow2u^2-12u+16=0\) \(\Leftrightarrow u^2-6u+8=0\) \(\Leftrightarrow\left(u-2\right)\left(u-4\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}u=2\left(loại\right)\\u=4\left(nhận\right)\end{matrix}\right.\). 

\(\Rightarrow v=6-u=2\), suy ra \(\left\{{}\begin{matrix}x+\dfrac{2}{x}=4\\y+\dfrac{1}{y}=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\pm\sqrt{2}\\y=1\end{matrix}\right.\) (nhận).

 Vậy hpt đã cho có các nghiệm \(\left(x;y\right)\in\left\{\left(2-\sqrt{2};1\right);\left(2+\sqrt{2};1\right)\right\}\)

Ta có: \(8-y^2=\left|xy-4\right|\ge0\Rightarrow y^2\le8\) (1)

Xét phương trình: \(x^2+2=xy\Leftrightarrow x^2-xy+2=0\)

\(\Leftrightarrow x^2-xy+\dfrac{y^2}{4}-\dfrac{y^2}{4}+2=0\)

\(\Leftrightarrow\dfrac{y^2}{4}-2=\left(x-\dfrac{y}{2}\right)^2\ge0\Rightarrow y^2\ge8\) (2)

Từ (1); (2) \(\Rightarrow\left\{{}\begin{matrix}y^2\ge8\\y^2\le8\end{matrix}\right.\) \(\Rightarrow y^2=8\Rightarrow y=...\)

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