已知x2分之(根号5)-1 y=2分之(根号5+1) 求（1）2x2-xy+2y2 （2）x2-y2-xy

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已知x=2分之(根号5-1) y=2分之(根号5+1)
x+y=根号5,x-y=-1,xy=1
（1）2x^2-xy+2y^2
=2(x^2+y^2)-xy
=2(x^2+y^2+2xy-2xy)-xy
=2(x^2+2xy+y^2)-4xy-xy
=2(x+y)^2-5xy
=2*(根号5)^2-5*1
=10-5
=5
（2）x^2-y^2-xy
=(x+y)(x-y)-xy
=根号5*(-1)-1
=-根号5 -1

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