~ Bạn tham khảo ~
Ta có :
`|x+1| + (y-2)^2=1`
Mà `|x+1|>=0;(y-2)^2>=0`
\(⇒\left[ \begin{array}{l}\begin{cases} |x+1|=0\\(y-2)^2=1 \end{cases}\\\begin{cases} |x-1|=1\\(y-2)^2=0 \end{cases}\end{array} \right.\)
\(⇒\left[ \begin{array}{l}\begin{cases} x+1=0\\\left[ \begin{array}{l}y-2=1\\y-2=-1\end{array} \right. \end{cases}\\\begin{cases} \left[ \begin{array}{l}x-1=1\\x-1=-1\end{array} \right.\\y-2=0 \end{cases}\end{array} \right.\) \(\)
\(⇒\left[ \begin{array}{l}\begin{cases} x=-1\\\left[ \begin{array}{l}y=3\\y=1\end{array} \right. \end{cases}\\\begin{cases} \left[ \begin{array}{l}x=2\\x=0\end{array} \right.\\y=2 \end{cases}\end{array} \right.\) \(\)
Vậy `...`
