已知x+y=1,x^2+y^2=2求.x^7+y^7的值

因为x+y=1,x^2+y^2=2,
又因为x^2+y^2=(x+y)^2-2xy=1^2-2xy=2,
所以2xy=-1,
所以xy=-1/2,
所以x^3+y^3=(x+y)(x^2-xy+y^2)=1*(2-xy)=1*[2-(-1/2)]=2.5,
x^4+y^4=(x^2+y^2)^2-2x^2y^2=2^2-2*(-1/2)^2=3.5,
所以x^7+y^7=(x^3+y^3)(x^4+y^4)-x^3y^4-x^4y^3
=(x^3+y^3)(x^4+y^4)-(xy)^3(x+y)
=2.5*3.5-(-0.5)^3*1
=8.75-(-0.125)
=8.875.

x^2+y^2=(x+y)^2-2xy
因为x+y=1,x^2+y^2=2
所以xy=(1-2)/2=-1/2

x^3+y^3=(x+y)(x^2-xy+y^2)=1*(2-(-1/2))=5/2

x^4+y^4=(x^2)^2+(y^2)^2=(x^2+y^2)^2-2(x^2)(y^2)=2^2-2(-1/2)^2=4-(1/2)=7/2

x^6+y^6=(x^3)2+(y^3)^2=(x^3+y^3)^2-2(xy)^3=(5/2)^2-2(-1/2)^3=25/4+1/4=13/2

x^8+y^8=(x^4)^2+(y^4)^2=(x^4+y^4)^2-2(xy)^4=(7/2)^2-2*(-1/2)^4=49/4-1/8=97/8

x^9+y^9=(x^3)^3+(y^3)^3=(x^3+y^3)[(x^3)^2-x^3y^3+(y^3)^2]==(x^3+y^3)[x^6-(xy)^3+y^6]=5/2*[13/2-(-1/2)^3]=5/2*(53/8)=265/16

x^7+y^7=x^(8-1)+y^(8-1)=(x^8+y^8)[(1/x)+(1/y)]-[(x^9+y^9)/xy]=(97/8)[(x+y)/xy])]-[(x^9+y^9)/xy]=(97/8)*[1/(-1/2)]*[(265/16