Question

\(\left\{{}\begin{matrix}x^3=2x+y\\y^3=2y+x\end{matrix}\right.\)

Answer 1

Lần lượt cộng vế với vế và trừ vế cho vế của 2 pt ta được:

\(\left\{{}\begin{matrix}x^3+y^3=3x+3y\\x^3-y^3=x-y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)\left(x^2-xy+y^2-3\right)=0\\\left(x-y\right)\left(x^2+xy+y^2-1\right)=0\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x+y=0\\x-y=0\end{matrix}\right.\) \(\Rightarrow x=y=0\)

Th2: \(\left\{{}\begin{matrix}x+y=0\\x^2+xy+y^2-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=-x\\x^2-x^2+x^2-1=0\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(-1;1\right);\left(1;-1\right)\)

TH3: \(\left\{{}\begin{matrix}x^2-xy+y^2-3=0\\x-y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2-x^2+y^2-3=0\\x=y\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(\sqrt{3};\sqrt{3}\right);\left(-\sqrt{3};-\sqrt{3}\right)\)

TH4: \(\left\{{}\begin{matrix}x^2-xy+y^2-3=0\\x^2+xy+y^2-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-3xy-3=0\\\left(x+y\right)^2-xy-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-3xy-3=0\\2xy+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy=-1\\x+y=0\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(-1;1\right);\left(1;-1\right)\)

Answer 2

Lấy PT(1) trừ PT(2) ta có:
${}^{}-{}^{}=(2x+y)-(2y+x)=x-y$

$\iff (x-y)({}^{}+xy+{}^{})-(x-y)=0$

$\iff (x-y)({}^{}+xy+{}^{}-1)=0$

Khi đó ta xét 2TH sau:

TH1: $x-y=0\iff x=y$

Thay vào PT ban đầu:

${}^{}=2x+y=2x+x=3x$

Answer 3

\(\Leftrightarrow\left\{{}\begin{matrix}x^3-2x-y=0\\y^3=2y+x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^3-2x-y=0\\y^3-2y-x=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}-x^3+2y^3-3y=0\\y^3-2y-x=0\end{matrix}\right.\)