pt :
\(2x^6-2x^3y+y^2=320\Leftrightarrow x^6+\left(x^6-2x^3y+y^2\right)=320\)
\(\Leftrightarrow x^6+\left(x^3-y\right)^2=320\)
=> \(x^6\le320\Leftrightarrow-2\le x\le2\)
TH1: Nếu \(x=-2\Rightarrow x^6=64\Rightarrow\left(x^3-y\right)^2=320-64=256\Rightarrow\orbr{\begin{cases}x^3-y=-16\\x^3-y=16\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}y=x^3+16=\left(-2\right)^3+16=8\\y=x^3-16=\left(-2\right)^3-16=-24\end{cases}}\)
TH2: Nếu \(x=2\Rightarrow x^6=64\Rightarrow\left(x^3-y\right)^2=320-64=256\Rightarrow\orbr{\begin{cases}x^3-y=-16\\x^3-y=16\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}y=x^3+16=2^3+16=24\\y=x^3-16=2^3-16=-8\end{cases}}\)
TH3: Nếu \(\orbr{\begin{cases}x=-1\\x=1\end{cases}}\Rightarrow x^6=1\Rightarrow\left(x^3-y\right)^2=320-1=319\) (vô nghiệm nguyên)
TH4: Nếu \(x=0\Rightarrow x^6=0\Rightarrow\left(x^3-y\right)^2=320\)(vô nghiệm nguyên)
Vậy pt có nghiệm (x,y)=...
