Question

[ILoveMath]

\(Ghpt:\left\{{}\begin{matrix}x-\dfrac{1}{x^3}=y-\dfrac{1}{y^3}\\\left(x-4y\right)\left(2x-y+4\right)=-36\end{matrix}\right.\)

Answer 1

[Nguyễn Văn A]

\(\left\{{}\begin{matrix}x-\dfrac{1}{x^3}=y-\dfrac{1}{y^3}\left(1\right)\\\left(x-4y\right)\left(2x-y+4\right)=-36\left(2\right)\end{matrix}\right.\)

\(Đk:\left\{{}\begin{matrix}x,y\ne0\\x\ne4y\\2x\ne y-4\end{matrix}\right.\)

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

\(\Rightarrow x-y+\dfrac{1}{y^3}-\dfrac{1}{x^3}=0\)

\(\Rightarrow x-y+\dfrac{x^3-y^3}{x^3y^3}=0\)

\(\Rightarrow x-y+\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{x^3y^3}=0\)

\(\Rightarrow\left(x-y\right).\dfrac{x^2+xy+y^2+x^3y^3}{x^3y^3}=0\)

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

Với \(x=y\) . Thay vào (2) ta được:

\(\left(x-4x\right)\left(2x-x+4\right)=-36\)

\(\Leftrightarrow-3x.\left(x+4\right)=-36\)

\(\Leftrightarrow x\left(x+4\right)=12\)

\(\Leftrightarrow x^2+4x-12=0\)

\(\Leftrightarrow\left(x+2\right)^2-16=0\)

\(\Leftrightarrow\left(x+6\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\Rightarrow y=2\\x=-6\Rightarrow y=-6\end{matrix}\right.\)

Với \(x^2+xy+y^2+x^3y^3=0\) . Ta sẽ chứng minh trường hợp này vô nghiệm.

Có: \(\left(x+y\right)^2+x^3y^3-xy=0\)

\(\Rightarrow\left(x+y\right)^2+xy\left(xy+1\right)\left(xy-1\right)=0\left(3\right)\)

Với \(xy>1\Rightarrow VT\left(3\right)>0\Rightarrow ptvn\)

Với \(xy=1\Rightarrow\left(x+y\right)^2=0\Rightarrow x=-y\)

\(\Rightarrow x^2=-1\Rightarrow ptvn\)

Với \(1>xy\ge0\Rightarrow xy\left(xy+1\right)\left(xy-1\right)\le0\) (có thể xảy ra).

Với \(0>xy>-1\Rightarrow VT\left(3\right)>0\Rightarrow ptvn\)

Với \(xy< -1\Rightarrow xy\left(xy-1\right)\left(xy+1\right)\le0\) (có thể xảy ra).

Vì \(x,y\ne0\) nên ta có: \(\left[{}\begin{matrix}1>xy>0\\xy< -1\end{matrix}\right.\left('\right)\)

\(\left(2\right)\Rightarrow2x^2-xy+4x-8xy+4y^2-16y=-36\)

\(\Rightarrow2x^2+4x+4y^2-16y+36=9xy\)

\(\Rightarrow2\left(x^2+2x+1\right)+4\left(y^2-4y+4\right)+18=9xy\)

\(\Rightarrow2\left(x+1\right)^2+4\left(y-2\right)^2+18=9xy>18\)

\(\Rightarrow xy>2\left(''\right)\)

Từ \(\left('\right),\left(''\right)\) suy ra hệ vô nghiệm.

Vậy hệ phương trình đã cho có nghiệm \(\left(x,y\right)\in\left\{\left(2;2\right),\left(-6;-6\right)\right\}\)