Question

设x+y=4,x的两次方+y的两次方=14,求x的5次方+y的5次方

Answer 1

x^2+y^2=(x+y)^2 -2xy=16-2xy=14 =>xy=1

=> x^4+y^4=(x^2+y^2)^2 -2x^2y^2=196-2=194

=> x^5+y^5=(x+y)(x^4-x^3y+x^2y^2-xy^3+y^4)=4(194-xy[x^2+y^2]+1)=4(194-14+1)=724