Question

x+y=1 x的2次方加y的2次方等于2 ,求x的7次方加y的7次方

x+y=1 x的2次方加y的2次方等于2 ,求x的7次方加y的7次方

Answer 1

x+y=1 x的2次方加y的2次方等于2 ,求x的7次方加y的7次方
(x+y)的2次方=1
因为x的2次方加y的2次方等于2
所以2xy=-1
x=(1-根号3)/2，y=(1+根号3)/2
或x=(1+根号3)/2，y=(1-根号3)/2
x的7次方加y的7次方
=(x+y)(x^6-x^5y+x^4*y^2-x^3*y^3+x^2*y^4-xy^5+y^6)
=x^6-x^5y+x^4*y^2-x^3*y^3+x^2*y^4-xy^5+y^6
=x^6+y^6-xy（x^4+y^4）+2x^2*y^2*(x^2+y^2)+1/8
=(x^2+y^2)(x^4-x^2y^2+y^4)+（x^4+y^4）/2+1+1/8
=2(x^4-1/4+y^4)+（x^4+y^4）/2+9/8
=5（x^4+y^4）/2+5/8
=5[（x^2+y^2）^2-2x^2y^2]/2+5/8
=5*(4-1/2)/2+5/8
=35/4+5/8
=75/8
=9.375

Answer 2

x^7+y^7＝(x+y)(x^6-x^5y+x^4y^2-x^3y^3+x^2y^4-xy^5+y^6)
＝x^6+y^6-(x^5y+xy^5)+(x^4y^2+x^2y^4)-x^3y^3
＝(x^2+y^2)(x^4-x^2y^2+y^4)-xy(x^4+y^4)+x^2y^2(x^2+y^2)-x^3y^3
＝2[(x^2+y^2)^2-3x^2y^2]+2x^2y^2-(xy)^3
＝2(4-3x^2y^2)+2x^2y^2-(xy)^3
＝8-6x^2y^2-(xy)^3
xy＝1/2［(x+y)^2-(x^2+y^2)］＝-1/2
所以原式＝8-3/2+1/8＝65/8-12/8
＝53/8