Question

Cho \(\sqrt{y-1}+\sqrt{|x-y+3|+3}=\sqrt{3}\)

Tìm x,y

Answer 1

Ta thấy: \(\left|x-y+3\right|\ge0\forall x,y\)

\(\Rightarrow\left|x-y+3\right|+3\ge3\forall x,y\)

\(\Rightarrow\sqrt{\left|x-y+3\right|+3}\ge\sqrt{3}\forall x,y\)

Lại có: \(\sqrt{y-1}\ge0\forall x;y\)

\(\Rightarrow VT=\sqrt{y-1}+\sqrt{\left|x-y+3\right|+3}\ge\sqrt{3}=VP\)

Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}\sqrt{y-1}=0\\\sqrt{\left|x-y+3\right|+3}=\sqrt{3}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}y-1=0\\\left|x-y+3\right|+3=3\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}y=1\\x-y+3=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-2\\y=1\end{matrix}\right.\)