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Viết lại (2)
\(xy\left(x^2+y^2\right)+2-\left(x+y\right)^2=0\)
\(\Leftrightarrow xy\left(x+y\right)^2-2x^2y^2+2-\left(x+y\right)^2=0\)
\(\Leftrightarrow\left(x+y\right)^2\left(xy-1\right)-2\left(x^2y^2-1\right)=0\)
\(\Leftrightarrow\left(xy-1\right)\left[\left(x+y\right)^2-2\left(xy+1\right)\right]=0\)
\(\Leftrightarrow\left(xy-1\right)\left(x^2+y^2-2\right)=0\)
- TH1: \(xy=1\)
\(\left(1\right)\Rightarrow5x-4y+3y^3-2\left(x+y\right)=0\)
\(\Leftrightarrow3x-6y+3y^3=0\)
\(\Leftrightarrow\dfrac{3}{y}-6y+3y^3=0\)
Đến đây dễ rồi nhé.
- TH2: \(x^2+y^2=2\)
\(\left(1\right)\Rightarrow5x^2y-4xy^2+3y^3-\left(x^2+y^2\right)\left(x+y\right)=0\)
\(\Leftrightarrow-x^3+2y^3+4x^2y-5xy^2=0\)
\(\Leftrightarrow\left(x-y\right)^2\left(x-2y\right)=0\)
Đến đây dễ rồi nhé.
![](https://rs.olm.vn/images/avt/0.png?1311)
b)\(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\)
\(\Rightarrow\left(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}\right)^2=\left(3\left(x+y\right)\right)^2\)
\(\Leftrightarrow\sqrt{\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)}=x^2+7xy+y^2\)
\(\Rightarrow\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)=\left(x^2+7xy+y^2\right)^2\)
\(\Leftrightarrow9\left(x-y\right)^2\left(x+y\right)^2=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)
\(\rightarrow\left(x;y\right)\in\left\{\left(0;0\right),\left(1;1\right)\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(\left\{{}\begin{matrix}x^2+y^4+xy=2xy^2+7\\xy^3-x^2y+4xy+11x=28+11y^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-y^2\right)^2+xy-7=0\\\left(x^{ }-y^2\right)\left(11-xy\right)+4\left(xy-7\right)=0\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x-y^2=a\\xy-7=b\end{matrix}\right.\) hệ trở thành \(\left\{{}\begin{matrix}a^2+b=0\\a\left(4-b\right)+4b=0\end{matrix}\right.\)\(\Rightarrow a\left(4+a^2\right)-4a^2=0\Leftrightarrow a\left(a^2-4a+4\right)=0\Leftrightarrow a\left(a-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}a=0;b=0\\a=2;b=-4\end{matrix}\right.\)
Giải từng trường hợp rồi kết hợp nghiệm
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x^2-2x\right)-\left(y^2-4y\right)=1\\\left(x^2-2x\right)^2+2=y\left(x-2\right)x\left(y-4\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x^2-2x\right)-\left(y^2-4y\right)=1\\\left(x^2-2x\right)^2+2=\left(x^2-2x\right)\left(y^2-4y\right)\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x^2-2x=u\\y^2-4y=v\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2u-v=1\\u^2+2=uv\end{matrix}\right.\) \(\Rightarrow u^2+2=u\left(2u-1\right)\)
\(\Leftrightarrow u^2-u-2=0\Leftrightarrow...\)
Dễ thấy x=0 không là nghiệm, xét x khác 0, đặt y=kx thì
\(\begin{cases} x^2(6-k-2k^2)=56\\ x^2(5-k-k^2)=49 \end{cases} \Rightarrow 49(6-k-2k^2)=56(5-k-k^2)\)
Từ đó tìm ra k và x,y nhé!