Question

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

Answer 1

ĐKXĐ: \(-1\le x;y\le1\)

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

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

\(\Rightarrow2x-2\sqrt{3}y+2=0\Rightarrow x=\sqrt{3}y-1\)

\(\Rightarrow\sqrt{1-y^2}=2-\sqrt{3}y\)

\(\Leftrightarrow1-y^2=3y^2-4\sqrt{3}y+4\)

\(\Leftrightarrow4y^2-4\sqrt{3}y+3=0\) \(\Rightarrow y=\frac{\sqrt{3}}{2}\Rightarrow x=\frac{1}{2}\)