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Lời giải:
Lấy $x.\text{PT(1)}+y.\text{PT(2)}$ thu được:
$3x^3+y^3=-2x^2y^2$
Lấy $x.\text{PT(1)}-y\text{PT(2)}$ thu được:
$3x^3-y^3=4xy$
$\Rightarrow y^3=-x^2y^2-2xy$
PT (2)$\Leftrightarrow 2x^2y+2y^2=-4x$
$\Leftrightarrow 2x^2y+y(xy^2+3x^2)=-4x$
$\Leftrightarrow x[2xy+y(y^2+3x)]=-4x$
$\Leftrightarrow x(y^3+5xy)=-4x$
$\Leftrightarrow x=0$ hoặc $y^3+5xy=-4$
Nếu $x=0$ thì dễ tìm $y=0$
Nếu $y^3+5xy=-4$
$\Leftrightarrow -x^2y^2-2xy+5xy=-4$
$\Leftrightarrow -(xy)^2+3xy+4=0$
$\Leftrightarrow (4-xy)(xy+1)=0$
$\Leftrightarrow xy=4$ hoặc $xy=-1$
Nếu $xy=4$ thì:
$y^3=-4-5xy=-24\Rightarrow y=\sqrt[3]{-24}$
$x^3=\frac{y^3+4xy}{3}=\frac{-8}{3}\Rightarrow x=\sqrt[3]{\frac{-8}{3}}$ (tm)
Nếu $xy=-1$ thì:
$y^3=-4-5xy=1\Rightarrow y=1$
$x^3=\frac{y^3+4xy}{3}=-1\Rightarrow x=-1$ (tm)
Vậy..........
\(1,\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\2y+10+y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{16}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}3x=1-2y\\1-2y+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\3y+6+2y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)
\(1,\Leftrightarrow\left\{{}\begin{matrix}x=2y+4\\-4y-8+5y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\cdot5+4=14\\y=5\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}5x-30+6x=3\\y=10-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=4-2y\\6y-12+y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{7}\\y=\dfrac{19}{7}\end{matrix}\right.\)
\(2x^2-\left(3y-3\right)x+y^2-2y+1=0\)
\(\Delta=\left(3y-3\right)^2-8\left(y^2-1y+1\right)=\left(y-1\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3y-3+y-1}{4}\\x=\dfrac{3y-3-y+1}{4}\end{matrix}\right.\)
\(\Rightarrow...\)
a) thay \(x^2y^2=2y^2-1\) vào PT (2):
\(\left(xy+1\right)\left(2y-x\right)=2x\left(2y^2-1\right)\)
\(\Leftrightarrow2xy^2-x^2y+2y-x=4xy^2-2x\)
\(\Leftrightarrow2xy^2-x+x^2y-2y=0\)
\(\Leftrightarrow\left(xy-1\right)\left(2y+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}xy=1\\x=-2y\end{matrix}\right.\)...
b)
\(\Leftrightarrow\left\{{}\begin{matrix}4x^2-6xy+2y^2=6\\x^2+2xy-2y^2=6\end{matrix}\right.\)
\(\Rightarrow3x^2-8xy+4y^2=0\)
\(\Rightarrow\left(3x-2y\right)\left(x-2y\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}y=\dfrac{3}{2}x\\y=\dfrac{1}{2}x\end{matrix}\right.\)
Thế vào pt đầu...
\(\left\{{}\begin{matrix}2x^2-3xy+y^2=3\\x^2+2xy-2y^2=6\end{matrix}\right.\)\(\left(1\right)\)\(\Leftrightarrow\left\{{}\begin{matrix}4x^2-6xy+2y^2=6\\x^2+2xy-2y^2=6\end{matrix}\right.\)
\(\Leftrightarrow3x^2-8xy+4y^2=0\)
\(\Leftrightarrow3x\left(x-2y\right)-2y\left(x-2y\right)=0\)
\(\Leftrightarrow\left(x-2y\right)\left(3x-2y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2y\\x=\dfrac{2y}{3}\end{matrix}\right.\)
Thay vào \(\left(1\right)\) ta được:
\(\Leftrightarrow\left[{}\begin{matrix}2.\left(2y\right)^2-3.2y.y+y^2=3\\2.\left(\dfrac{2y}{3}\right)^2-3.\dfrac{2y}{3}.y+y^2=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}y^2=1\\y^2=-27\left(VLý\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\end{matrix}\right.\)
Vậy ...