三道数学题(解出其中一道也行)1.已知x+y=5,x²+y²=13,求代数式x³y+2x²y²+xy³的值2.已知x²+2x+y²-6y+10=0,求x的y次方的值3.已知x(x-1)-(x²-y)+2=0,求代数式x²+y²/2-xy的值

问题描述:

三道数学题(解出其中一道也行)
1.已知x+y=5,x²+y²=13,求代数式x³y+2x²y²+xy³的值
2.已知x²+2x+y²-6y+10=0,求x的y次方的值
3.已知x(x-1)-(x²-y)+2=0,求代数式x²+y²/2-xy的值

1、x+y=5,x2+y2=13 x2+y2+2xy=25 13+2xy=25, xy=6
x3y+2x2y2+xy3=xy(x2+y2)+2x2y2=6*13+2*36=150
2、x2+2x+y2-6y+10=0 x2+2x+1+y2-6y+9=0 (x+1)^2+(y-3)^2=0
x=-1,y=3 x的y次方=(-1)^3=-1

1、x+y=5,x2+y2=13 x2+y2+2xy=25 13+2xy=25, xy=6
x3y+2x2y2+xy3=xy(x2+y2)+2x2y2=6*13+2*36=150
2、x2+2x+y2-6y+10=0 x2+2x+1+y2-6y+9=0 (x+1)^2+(y-3)^2=0
x=-1,y=3 x的y次方=(-1)^3=-1
3、x(x-1)-(x2-y)+2=0 y=x-2
x2+y2/2-xy=x2+y2/2-xy=(x-y)^2/2=2

1.解两个方程构成的方程组,有:
x=2,y=3

x=3,y=2
所以有:
(1)x³y+2x²y²+xy³=24+72+54=150
(2)x³y+2x²y²+xy³=54+72+24=150
也就是,结果为150.
解法二:x³y+2x²y²+xy³=xy(x+y)^2={[(x+y)^2-(x²+y²)]/2}(x+y)^2={[5^2-13]/2}*5^2=150
2.x²+2x+y²-6y+10=(x+1)^2+(y-3)^2=0
故,x=-1,y=3
有,x^y=-1
3.x(x-1)-(x²-y)+2=y-x+2=0
这说明:x-y=2
x²+y²/2-xy=(x-y)^2/2=2