Question

Giải hệ phương trình: \(\left\{{}\begin{matrix}xy\left(x+y\right)=2\\x^3+y^3+6=8x^2y^2\end{matrix}\right.\)

Answer 1

ĐK : \(x;y\ne0;x\ne-y\)\(\left\{{}\begin{matrix}xy\left(x+y\right)=2\\x^3+y^3+6=8x^2y^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)=2;x+y=\dfrac{2}{xy}\\x^3+y^3+3.xy\left(x+y\right)=8x^2y^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)=2;x+y=\dfrac{2}{xy}\\\left(x+y\right)^3=8x^2y^2\end{matrix}\right.\)

Lại có : \(\left(x+y\right)^3=8x^2y^2\)

\(\Leftrightarrow\left(\dfrac{2}{xy}\right)^3=8x^2y^2\)

\(\Leftrightarrow\dfrac{8}{x^3y^3}=8x^2y^2\)

\(\Leftrightarrow\dfrac{1}{x^3y^3}=x^2y^2\)

\(\Leftrightarrow1=\left(xy\right)^5\)

\(\Leftrightarrow xy=1\)

Do xy = 1 \(\Rightarrow x+y=\dfrac{2}{1}=2\)

Ta lại có :

\(x^3+y^3+6=8x^2y^2\)

\(\Leftrightarrow2\left(x^2+y^2-xy\right)+6=8\)

\(\Leftrightarrow x^2+y^2-xy=1\)

\(\Leftrightarrow x^2+y^2=2xy\)

\(\Leftrightarrow\left(x-y\right)^2=0\Leftrightarrow x=y\)

Mà \(x+y=2\)

\(\Rightarrow x=y=1\)

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