Question

giai hpt

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

Answer 1

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

ĐKXĐ: x;y≥0

Ta thấy: \(x+3\ne x\Rightarrow\sqrt{x+3}-\sqrt{x}\ne0\)

=>(1)<=>\(\frac{3\sqrt{y}}{\sqrt{x+3}-\sqrt{x}}=3\Leftrightarrow\sqrt{x}+\sqrt{y}=\sqrt{x+3}\). Thay vào (2):

\(\sqrt{x+3}=x+1\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x^2+x-2=0\end{matrix}\right.\Leftrightarrow x=1\) (tm đkxđ)

Thay vào (1) => y=1 (tm đkxđ)

Vậy hệ có nghiệm (x;y)\(\in\){(1;1)}