题目已知xxx，yyy满足方程组{3x2-2xy+12y2=47,2x2+xy+8y2=36.\left\{\begin{matrix}3x^{2}\text{-}2\text{xy}\text{+}12y^{2}\text{=}47\text{,}\\ {2x}^{2}\text{+}\text{xy}\text{+}8y^{2}\text{=}36\text{.}\\\end{matrix}\right.{3x2-2xy+12y2=47,2x2+xy+8y2=36.则x2+4y2x^{2}+4y^{2}x2+4y2的值为()(\quad)()A.111111B.171717C.141414D.161616知识点：解二元一次方程组——加减消元法章节：方程与不等式 / 二元一次方程组
