Đáp án :
$\begin{array}{l}\Rightarrow\left\{\begin{matrix}\left[ \begin{array}{l}x=0\\x=-2\end{array} \right.& \\\left[ \begin{array}{l}y=3\\y=-3\end{array} \right.& \end{matrix}\right.\end{array}$
Giải thích các bước giải :
`|x^2+2x|+|y^2-9|=0` (*)
Vì `|x^2+2x| ≥ 0; |y^2-9| ≥ 0`
Để xảy ra (*)
$\begin{array}{l}\Rightarrow\left\{\begin{matrix}\ |x^2+2x|=0& \\ |y^2-9|=0 \end{matrix}\right.\end{array}$
$\begin{array}{l}\Rightarrow\left\{\begin{matrix}\ x^2+2x=0& \\ y^2-9=0 \end{matrix}\right.\end{array}$
$\begin{array}{l}\Rightarrow\left\{\begin{matrix}\ x(x+2)=0& \\ (y-3)(y+3)=0 \end{matrix}\right.\end{array}$
$\begin{array}{l}\Rightarrow\left\{\begin{matrix}\left[ \begin{array}{l}x=0\\x+2=0\end{array} \right. & \\\left[ \begin{array}{l}y-3=0\\y+3=0\end{array} \right.& \end{matrix}\right.\Rightarrow\left\{\begin{matrix}\left[ \begin{array}{l}x=0\\x=-2\end{array} \right.& \\\left[ \begin{array}{l}y=3\\y=-3\end{array} \right.& \end{matrix}\right.\end{array}$
Vậy $\begin{array}{l}\Rightarrow\left\{\begin{matrix}\left[ \begin{array}{l}x=0\\x=-2\end{array} \right.& \\\left[ \begin{array}{l}y=3\\y=-3\end{array} \right.& \end{matrix}\right.\end{array}$
