Question

Giải hệ phương trình: \(\left\{{}\begin{matrix}\left|x+5\right|-\frac{2}{\sqrt{y}-2}=4\\\left|x+5\right|+\frac{1}{\sqrt{y}-2}=3\end{matrix}\right.\)

Answer 1

ĐK: \(y\ge0,y\ne4\)

Đặt \(a=\left|x+5\right|;b=\frac{1}{\sqrt{y}-2}\left(a\ge0\right)\)

\(\Rightarrow\left\{{}\begin{matrix}a-2b=4\\a+b=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=\frac{10}{3}\\b=\frac{-1}{3}\end{matrix}\right.\)(TM)\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x+5=\frac{10}{3}\\x+5=\frac{-10}{3}\end{matrix}\right.\\\sqrt{y}-2=-3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x+5=\frac{10}{3}\\x+5=\frac{-10}{3}\end{matrix}\right.\\\sqrt{y}=-1\left(vl\right)\end{matrix}\right.\)

Vậy hpt vô nghiệm.

Answer 2

theo bài ra ta có

\(\frac{3}{\sqrt{y}-2}=-1\\ \Leftrightarrow-3=\sqrt{y}-2\\ \Leftrightarrow\sqrt{y}=-1\)

k\(^o\) có y t/m

phương trình vô ng\(^o\)