ĐKXĐ: ...
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{\left(x^2y+1\right)\left(3-x^2y\right)}=2x^6+y^4-x^4\\0=x^6+2x^3y-x^4-1-\sqrt{1+\left(x-y\right)^2}\end{matrix}\right.\)
Trừ vế cho vế:
\(\sqrt{\left(x^2y+1\right)\left(3-x^2y\right)}=x^6-2x^3y+y^4+1+\sqrt{1+\left(x-y\right)^2}\)
\(\Leftrightarrow\sqrt{\left(x^2y+1\right)\left(3-x^2y\right)}=\left(x^3-y^2\right)^2+1+\sqrt{1+\left(x-y\right)^2}\)
\(VP=\left(x^3-y^2\right)^2+1+\sqrt{1+\left(x-y\right)^2}\ge0+1+\sqrt{1}=2\)
\(VT=\sqrt{\left(x^2y+1\right)\left(3-x^2y\right)}\le\frac{1}{2}\left(x^2y+1+3-x^2y\right)=2\)
Dấu "=" xảy ra khi và chỉ khi \(x=y=1\)
