Question

若已知x+y=1,x^2+y^2=2,求x^4+y^4和x^4-y^4的值

Answer 1

(x+y)^2=x^2+y^2+2xy=1 (x-y)^2=x^2+y^2-2xy=2+1=3 x-y=±√3

2xy=-1 xy=-1/2 x^2y^2=1/4

(x^2+y^2)^2=x^4+y^4+2x^2y^2=4

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

x^4-y^4=(x^2+y^2)(x^2-y^2）=(x^2+y^2)(x+y)(x-y)=2*1*√3 =±2√3

肯定要有的,如果不要根号3就取近似值1.732代入吧