12．（5分）若$x$，$y$满足$x^{2}+y^{2}-xy=1$，则$($　　$)$
A．$x+y\leqslant 1$              B．$x+y\geqslant -2$              C．$x^{2}+y^{2}\leqslant 2$              D．$x^{2}+y^{2}\geqslant 1$
分析：方法一：原等式可化为，$(x-\dfrac{y}{2})^{2}+(\dfrac{\sqrt{3}}{2}y)^{2}=1$，进行三角代换，令$\left\{\begin{array}{l}{x-\dfrac{y}{2}=\cos \theta }\\ {\dfrac{\sqrt{3}}{2}y=\sin \theta }\end{array}\right.$，则$\left\{\begin{array}{l}{x=\dfrac{\sqrt{3}}{3}\sin \theta +\cos \theta }\\ {y=\dfrac{2\sqrt{3}}{3}\sin \theta }\end{array}\right.$，结合三角函数的性质分别求出$x+y$与$x^{2}+y^{2}$的取值范围即可．
方法二：由$x^{2}+y^{2}-xy=1$可得，$(x+y)^{2}=1+3xy\leqslant 1+3(\dfrac{x+y}{2})^{2}$，$x^{2}+y^{2}-1=xy\leqslant \dfrac{{x}^{2}+{y}^{2}}{2}$，分别求出$x+y$与$x^{2}+y^{2}$的取值范围即可．
解：方法一：由$x^{2}+y^{2}-xy=1$可得，$(x-\dfrac{y}{2})^{2}+(\dfrac{\sqrt{3}}{2}y)^{2}=1$，
令$\left\{\begin{array}{l}{x-\dfrac{y}{2}=\cos \theta }\\ {\dfrac{\sqrt{3}}{2}y=\sin \theta }\end{array}\right.$，则$\left\{\begin{array}{l}{x=\dfrac{\sqrt{3}}{3}\sin \theta +\cos \theta }\\ {y=\dfrac{2\sqrt{3}}{3}\sin \theta }\end{array}\right.$，
$\therefore x+y=\sqrt{3}\sin \theta +\cos \theta =2\sin (\theta +\dfrac{\pi }{6})\in [-2$，$2]$，故$A$错，$B$对，
$\because x^{2}+y^{2}=(\dfrac{\sqrt{3}}{3}\sin \theta +\cos \theta )^{2}+(\dfrac{2\sqrt{3}}{3}\sin \theta )^{2}=\dfrac{\sqrt{3}}{3}\sin 2\theta -\dfrac{1}{3}\cos 2\theta +\dfrac{4}{3}=\dfrac{2}{3}\sin (2\theta -\dfrac{\pi }{6})+\dfrac{4}{3}\in [\dfrac{2}{3}$，$2]$，
故$C$对，$D$错，
方法二：对于$A$，$B$，由$x^{2}+y^{2}-xy=1$可得，$(x+y)^{2}=1+3xy\leqslant 1+3(\dfrac{x+y}{2})^{2}$，即$\dfrac{1}{4}(x+y)^{2}\leqslant 1$，
$\therefore (x+y)^{2}\leqslant 4$，$\therefore -2\leqslant x+y\leqslant 2$，故$A$错，$B$对，
对于$C$，$D$，由$x^{2}+y^{2}-xy=1$得，$x^{2}+y^{2}-1=xy\leqslant \dfrac{{x}^{2}+{y}^{2}}{2}$，
$\therefore x^{2}+y^{2}\leqslant 2$，故$C$对；
$\because -xy\leqslant \dfrac{{x}^{2}+{y}^{2}}{2}$，$\therefore 1=x^{2}+y^{2}-xy\leqslant x^{2}+y^{2}+\dfrac{{x}^{2}+{y}^{2}}{2}=\dfrac{3({x}^{2}+{y}^{2})}{2}$，
$\therefore$${x}^{2}+{y}^{2}\geqslant \dfrac{2}{3}$，故$D$错误．
故选：$BC$．
点评：本题主要考查了三角代换求最值，考查了三角函数的性质，同时考查了学生分析问题，转化问题的能力，属于中档题．