Question

(3xy+2)的平方+7-x-y的绝对值=0

若(3xy+2)的二次方加7-x-y的绝对值等于0,求代数式(5xy+10y)-[-5x-(4x
-2y+3x)]的值.

Answer 1

(3xy+2)^2+|7-x-y|=0,
∴xy=-2/3,x+y=7,
(5xy+10y)-[-5x-(4x+2y+3x)](改题了)
=5xy+10y-(-5x-7x-2y)
=5xy+10y+12x+2y
=5xy+12(x+y)
=-10/3+84
=242/3.