Question

已知(x+1)的平方+y-1的绝对值=0 求2(xy-5xy的平方)-(3xy的平方-xy)的值

Answer 1

(x+1)^2+|y-1|=0
(x+1)^2>=0
|y-1|>=0
x+1=0,y-1=0
x=-1,y=1
2(xy-5xy的平方)-(3xy的平方-xy)
=2xy-10xy^2-3xy^2+xy
=3xy-13xy^2
=-3+13
=10