Question

tìm x,y biết rằng x+y+12=\(4\sqrt{x}-6\sqrt{y-1}\)

Answer 1

\(x+y+12=4\sqrt{x}-6\sqrt{y-1}\) (ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\y\ge1\end{matrix}\right.\))

\(\Leftrightarrow\left(x-4\sqrt{x}+4\right)+\left[\left(y-1\right)+6\sqrt{y-1}+9\right]=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+\left(\sqrt{y-1}+3\right)^2=0\)

Vì \(\left\{{}\begin{matrix}\left(\sqrt{x}-2\right)^2\ge0\\\left(\sqrt{y-1}+3\right)^2\ge9\end{matrix}\right.\Rightarrow VT\ge9\)

Vậy pt vô nghiệm.