Question

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

Answer 1

\(x;y\ge-3;xy\ne0\)

\(\frac{1}{x}+\frac{1}{y}=2\Leftrightarrow x+y=2xy\)

\(\sqrt{x+3}+\sqrt{y+3}=4\Leftrightarrow x+y+6+2\sqrt{xy+3\left(x+y\right)+9}=16\)

\(\Leftrightarrow2xy+6+2\sqrt{7xy+9}=16\Leftrightarrow\sqrt{7xy+9}=5-xy\) (\(xy\le5\))

\(\Leftrightarrow7xy+9=\left(5-xy\right)^2\Leftrightarrow\left(xy\right)^2-17xy+16=0\)

\(\Rightarrow\left[{}\begin{matrix}xy=16>5\left(l\right)\\xy=1\Rightarrow x+y=2xy=2\end{matrix}\right.\)

Theo Viet đảo, x; y là nghiệm của:

\(t^2-2t+1=0\Rightarrow t=1\Rightarrow x=y=1\)