Question

于完全平方公式.1.已知x^2+y^2=2,求x^2+y^2-4xy-2(x-y)^2的值2.若a-1\a=4,求1\a

Answer 1

1、x^2+y^2-4xy-2(x-y)^2

= x^2+y^2-2xy-2xy-2(x-y)^2

= (x-y)^2-2xy-2(x-y)^2

= -(x-y)^2-2xy

= -x^2+ 2xy -y^2-2xy

= -x^2 -y^2

= -(x^2+y^2)

= -2

2、

因为a-1a=4

所以(a-1a)^2=16

即 a^2-2+1a^2=16

所以a^2+1a^2=18

3、(x+y)^2-(x-y)^2=4a^2-4b^2

化简后得到：4xy=4a^2-4b^2

即：xy=a^2-b^2

总结：多利用平方的性质