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