Question

giup minh cau nay voi

 

Answer 1

Đặt \(\left\{{}\begin{matrix}a=\left|x-1\right|\\b=\sqrt{y+2}\left(b>0\right)\end{matrix}\right.\)

Hệ phương trình trở thành:

 \(\left\{{}\begin{matrix}2a-b=4\\a+3b=9\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=2a-4\\a+3.\left(2a-4\right)=9\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=2a-4\\7a-12=9\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=2a-4\\a=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=3\\b=2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left|x-1\right|=3\\\sqrt{y+2}=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\\y=2\end{matrix}\right.\)

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