Vì \(\left|x^2+2x\right|\ge0\) và \(\left|y^2-9\right|\ge0\)
=> Dấu = xảy ra khi : \(\hept{\begin{cases}x^2-2x=0\\y^2-9=0\end{cases}}\) => \(\hept{\begin{cases}x\left(x+2\right)=0\\\left(y-3\right)\left(y+3\right)=0\end{cases}}\) => \(\hept{\begin{cases}x=\left\{0;2\right\}\\y=\left\{3;-3\right\}\end{cases}}\)
để \(\left|x^2+2x\right|+\left|y^2-9\right|=0\)
\(\Rightarrow\left|x^2+2x\right|=\left|y^2-9\right|=0\)
\(\Rightarrow\left|x^2+2x\right|=0\) \(\Rightarrow y^2-9=0\)
\(x^2+2x=0\) \(y^2-9=0\)
\(x^2=-2x\) \(y^2=9\)
\(x^2:x=-2\) \(y^2=3^2=\left(-3\right)^2\)
\(x=-2\) \(\Rightarrow y=3\)hoặc \(y=-3\)
vậy............
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