题目内容：

已知x=2-根号3分之2+根号3,y=2+根号3分之2-根号3,求值:2x^2-3xy+2y^2

优质解答

x=(2+√3)/(2-√3) 分母有理化
=(2+√3)²/[(2+√3)(2-√3)]
=(7+4√3)/(4-3)
=7+4√3
y=(2-√3)/(2+√3) 分母有理化
=(2-√3)²/[(2+√3)(2-√3)]
=(7-4√3)/(4-3)
=7-4√3
2x²-3xy+2y²
=(2x²-4xy+2y²)+xy
=2(x²-2xy+y²)+xy
=2(x-y)²+xy
=2[ (7+4√3)-(7-4√3) ]² +(7+4√3)(7-4√3)
=2×(8√3)² +49-48
=2×192 +1
=385