Tích chéo 2 hpt ta có:
\(6x^4-6y^4=15 x^3y-15y^3x\)
<=>\(6x^4-6y^4-15x^3y+15y^3x=0\)
<=> \(6(x^2-y^2)(x^2+y^2)-15xy (x^2-y^2)=0\)
<=>\((x^2-y^2)(6x^2+6y^2+15)=0\)
=> x2=y2
=> x=y hoặc x=-y
(*)x=y=>vô nghiệm
(*)x=-y=> vô no
Vậy hpt vô nghiệm
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+y^2\right)\left(x^2-y^2\right)=15\\x^2-y^2=\dfrac{6}{xy}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+y^2\right).\dfrac{6}{xy}=15\\x^2-y^2=\dfrac{6}{xy}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x^2+6y^2=15xy\\x^2-y^2=\dfrac{6}{xy}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(3x-3y\right)\left(2x-y\right)=0\\x^2-y^2=\dfrac{6}{xy}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=y\\x=\dfrac{y}{2}\end{matrix}\right.\\x^2-y^2=\dfrac{6}{xy}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=y\\x^2-y^2=\dfrac{6}{xy}\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{y}{2}\\x^2-y^2=\dfrac{6}{xy}\end{matrix}\right.\end{matrix}\right.\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=y\\x^2-x^2=\dfrac{6}{x^2}\left(vl\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{y}{2}\\\dfrac{y^2}{4}-y^2=\dfrac{12}{y^2}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{y}{2}\\\dfrac{3y^2}{4}+\dfrac{12}{y^2}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{y}{2}\\3y^4\text{+48}=0\end{matrix}\right.\left(vl\right)\)
vậy hpt không có nghiệm với mọi x,y
