Question

求x^2+y^2的值

已知xy+x+y=71，x^2y+xy^2=1260,x,y为正数，求x^2+y^2的值

Answer 1

xy+x+y=71，
x^2y+xy^2=1260,x,y为正数
xy(x+y)=1260
xy,x+y是方程z²-71z+1260=0的两个根
解得：z1=35，z2=36
xy=35,x+y=36
或xy=36,xy=35

x^2+y^2=（x+y)²-2xy=35²-2*36=1153
或
x^2+y^2=（x+y)²-2xy=36²-2*35=1226

Answer 2

xy+x+y=71
x+y=71-xy
x^2y+xy^2=1260
xy(x+y)=1260
xy(71-xy)=1260
(xy)^2-71(xy)+1260=0
(xy-35)(xy-36)=0
xy=35 或 36
当xy=35时
x+y=36
(x+y)^2=1296
x^2+y^2=1296-2xy=1226

当xy=36时
x+y=35
(x+y)^2=1225
x^2+y^2=1225-2xy=1153

Answer 3

1.xy+x+y=71，
2.x^2y+xy^2=1260
设xy=A X+Y=B
3.由1.2.得A
4.由2.3.得B
5.X2+Y2可由A,B来表示
6.自己算吧，加油

Answer 4

1226