已知x+y=-1, xy=-5 求x五次方+y五次方的值 好的绝对追加!!!
发布网友
发布时间：2024-10-24 02:42
共5个回答
热心网友
时间：2024-11-06 20:04
关键是因式分解
（x+y）²=1
x²+2xy+y²=1
x²+y²=1-2xy=11
(x+y)^3=(x+y)(x²-xy+y²)=-(11+5)=-16
(x²+y²)(x^3+y^3)=x^5+x^3y²+x²y^3+y^5=1x-16=-16
x^5+y^5=-16-x^3y²-x^3y^3=-16-x²y²(x+y)=-16+x²y²=-16+(xy)²=-16+25=9
热心网友
时间：2024-11-06 20:04
（x+y）²=1
x²+2xy+y²=1
x²+y²=1-2xy=11
(x+y)^3=(x+y)(x²-xy+y²)=-(11+5)=-16
(x²+y²)(x^3+y^3)=x^5+x^3y²+x²y^3+y^5=1x-16=-16
x^5+y^5=-16-x^3y²-x^3y^3=-16-x²y²(x+y)=-16+x²y²=-16+(xy)²=-16+25=9
热心网友
时间：2024-11-06 20:08
x+y=-1.x^2+y^2+2xy=1.
x^2+y^2=11.
(x^2+y^2)(x+y)=x^3+y^3+xy（x+y）=11
x^3+y^3=6
(x^3+y^3)(x^2+y^2)=x^5+y^5+（xy）^2(x+y)=66
x^5+y^5=91.
热心网友
时间：2024-11-06 20:06
x+y=-1
xy=-5
x^2+2xy+y^2=1
x^2+y^2=1-2xy=1+10=11
两边平方
x^4+2x^2y^2+y^4=121
x^4+y^4=121-2(xy)^2=121-50=71
x^3+y^3=(x+y)(x^2-xy+y^2)=-1*(11+5)=-16
x^5+y^5=(x+y)(x^4+y^4)-(x^4y+xy^4)
=(x+y)(x^4+y^4)-xy(x^3+y^3)
=-1*71-(-5)*(-16)
=-151
热心网友
时间：2024-11-06 20:06
x+y=-1,xy=-5
则(x+y)²=1
x²+y²=1-2xy=11
x³+y³=(x+y)(x²-xy+y²)
=3[(x+y)²-3xy]
=3×(1-3*(-5))
=48
x的五次方+y的五次方
=(x²+y²)(x³+y³)-x²y³-x³y²
=11×48-(xy)²(x+y)
=11×48-(-5)^2*(-1)
=553
