已知(x+1)2+|y-1|=0,求2(xy-5xy2)-(3xy2-xy)的值.

问题描述:

已知(x+1)2+|y-1|=0,求2(xy-5xy2)-(3xy2-xy)的值.

2(xy-5xy2)-(3xy2-xy)
=(2xy-10xy2)-(3xy2-xy)
=2xy-10xy2-3xy2+xy
=(2xy+xy)+(-3xy2-10xy2
=3xy-13xy2
∵(x+1)2+|y-1|=0
∴(x+1)=0,y-1=0
∴x=-1,y=1.
∴当x=-1,y=1时,
3xy-13xy2=3×(-1)×1-13×(-1)×12
=-3+13
=10.
答:2(xy-5xy2)-(3xy2-xy)的值为10.
答案解析:因为平方与绝对值都是非负数,且(x+1)2+|y-1|=0,所以x+1=0,y-1=0,解得x,y的值.再运用整式的加减运算,去括号、合并同类项,然后代入求值即可.
考试点:整式的加减—化简求值;非负数的性质:绝对值;非负数的性质:偶次方.
知识点:整式的加减运算实际上就是去括号、合并同类项,这是各地中考的常考点.代入求值时要化简.