已知\(x^2+2y^2-\sqrt{3}xy=1(x,y\in\mathbf{R}),\)则\(x^2+y^2\)的最小值为_________\(.\)
解答：
法一：设\(x^2+y^2=t,\)则\(x^2+y^2=t(x^2+2y^2-\sqrt{3}xy)\).
\((1-t)x^2+\sqrt{3}txy+y^2-2ty=0,\)
\(\Delta =3t^2y^2-4(1-t)(y^2-2ty^2)\geq 0,\)即得\(t\in [\dfrac{2}{5},2].\)
故\((x^2+y^2)_\min=\dfrac{2}{5}.\)
评注：这个方法将\(x\)看作主元，构造一元二次方程利用根的存在性求出\(x^2+y^2\)的取值范围.
法二：三角代换\(,\)配方可得\((x-\dfrac{\sqrt{3}}{2}y)^2+\dfrac{5}{4}y^2=1,\)
设\(x-\dfrac{\sqrt{3}}{2}y=\cos{\theta},\dfrac{\sqrt{5}}{2}y=\sin{\theta}\)