Ta có : \(\left(x-0,2\right)^{10};\left(y+3,1\right)^{20}\ge0\) với mọi \(x,y\)

Mà \(\left(x-0,2\right)^{10}+\left(y+3,1\right)^{20}=0\)

\(< =>\hept{\begin{cases}\left(x-0,2\right)^{10}=0\\\left(y+3,1\right)^{20}=0\end{cases}< =>\hept{\begin{cases}x-0,2=0\\y+3,1=0\end{cases}< =>\hept{\begin{cases}x=0,2\\y=-3,1\end{cases}}}}\)

Vậy \(x=0,2;y=-3,1\)

VÌ  \(\left(x-0,2\right)^{10}\ge0;\left(y+3,1\right)^{10}\ge0mà\left(x-0,2\right)^{10}+\left(y+3,1\right)^{10}=0\Rightarrow x-0,2=0;y+3,1=0\)

\(\left(x-0,2\right)^{10}+\left(y+3,1\right)^{20}=0\)

\(\Rightarrow\orbr{\begin{cases}x-0,2=0\\y+3,1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0,2\\y=-3,1\end{cases}}\)

\(\left(x-0,2\right)^{10}+\left(y+3,1\right)^{20}=0\)
\(=>\left[{}\begin{matrix}\left(x-0,2\right)^{10}=0\\\left(y+3,1\right)^{20}=0\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x-0,2=0\\y+3,1=0\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=0,2\\y=-3,1\end{matrix}\right.\)
Vậy...

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x+y+xy+1=0 => y+x(y+1)+1=0 => (y+1)+x(y+1)=0 => (x+1)(y+1)=0 => x=-1 thì y bất kì còn y = -1 thì x bất kì 

\(x+y+x.y+1=0\)

\(x.1+x.y+y+1\) \(=0\)

\(x.\left(1+y\right)+\left(y+1\right)\) \(=0\)

\(\left(1+y\right).\left(x+1\right)=0\)

\(\Rightarrow1+y=0\) \(\Rightarrow\) \(y=-1\)

\(\Rightarrow\) \(x+1=0\) \(\Rightarrow\) \(x=-1\)