Question

Giải hệ phương trình:

a) \(\left\{{}\begin{matrix}\left|x+1\right|+\sqrt{y}=5\\\left(x^2+2x+1\right)y=36\end{matrix}\right.\)

Answer 1

Lời giải:
HPT \(\Leftrightarrow \left\{\begin{matrix} |x+1|+\sqrt{y}=5\\ (x+1)^2y=36\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} |x+1|+\sqrt{y}=5\\ |x+1|\sqrt{y}=6\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} |x+1|=5-\sqrt{y}\\ |x+1|\sqrt{y}=6\end{matrix}\right.\)

\(\Rightarrow (5-\sqrt{y})\sqrt{y}=6\)

\(\Leftrightarrow y-5\sqrt{y}+6=0\)

\(\Leftrightarrow (\sqrt{y}-2)(\sqrt{y}-3)=0\Rightarrow \left[\begin{matrix} y=4\\ y=9\end{matrix}\right.\)

Nếu \(y=4\Rightarrow |x+1|=5-\sqrt{y}=5-2=3\)

\(\Rightarrow x+1=\pm 3\Rightarrow x\in \left\{-4;2\right\}\)

Nếu \(y=9\Rightarrow |x+1|=5-\sqrt{y}=2\Rightarrow x+1=\pm 2\)

\(\Rightarrow x\in\left\{-3;1\right\}\)

Vậy \((x,y)=(-4,4); (2,4);(-3,9); (1,9)\)