Question

已知根号下（x+y-3）+(xy+x+y-2)的绝对值=0,求代数式y分之x^2+x分之y^2的值

Answer 1

根号下（x+y-3）+(xy+x+y-2)的绝对值=0

根据根号和绝对值的性质

x+y-3＝0

xy+x+y-2＝0,xy+1=0

即x+y=3,xy=-1

y分之x^2+x分之y^2

=(x^3+y^3)/(xy)

=(x+y)(x^2+y^2-xy)/(xy)

=(x+y)[(x+y)^2-3xy]/(xy)

=3*(9+3)/(-1)

=-36