Question

已知x+y=1，x的平方+y的平方=2，求x的6次方+y的6次方的值

Answer 1

x+y=1，x^2+y^2=2
x^2+y^2=(x+y)^2-2xy=2
xy=1/2[(x+y)^2-1]=1/2(4-1)=3/2
x^6+y^6=(x^2+y^2)^3-3x^4y^2-3x^2y^4
=(x^2+y^2)^3-3x^2y^2(x^2+y^2)
=[(x+y)^2-2xy]^3-3x^2y^2(x^2+y^2)
=[1-(2*3/2)]^2-3*(3/2)^2*2
=-9.5

Answer 2

解：∵x+y=1，x^2+y^2=2
∴xy=((x+y)^2-(x^2+y^2))/2=-1/2
故x^6+y^6=(x^2)^3+(y^2)^3
=(x^2+y^2)(x^4-(xy)^2+y^4)
=(x^2+y^2)((x^2+y^2)^2-3(xy)^2)
=2*(2^2-3*(-1/2)^2)
=13/2。