Question

已知x+y=5／2，x^2+y^2=17／4，则x^4-y^4=?

Answer 1

解：∵x+y=5/2
∴(x+y)²=25/4 ①
∵ x²+y²=17/4 ②
①-②得：
2xy=8/4
xy=1
(x-y)²=(x²+y²)-2xy=17/4-2=9/4
∴x-y=3/2 或 x-y=-3/2
x^4-y^4=(x²+y²)(x²-y²)
=(x²+y²)(x+y)(x-y)
=17/4×5/2×(±3/2)
=±255/16

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Answer 2

解；因为x+y=5/2
则x^2+2xy+y^2=25/4
而x^2+y^2=17／4
则2xy=2,即xy=1
化简x^4-y^4
=(x^2+y^2)(x^2-y^2)
=(x^2+y^2)(x+y)(x-y)
其中x-y=根号下（x-y)^2=根号下x^2-2xy+y^2=根号下9/4

=3/2或-3/2
代入得(x^2+y^2)(x+y)(x-y)=255/16或-255/16

Answer 3

正负255/16，步骤、y=5/2-x,带入x^2+y^2+17/4得x=2或1/2，y=1/2或2带入得正负255/16.