试题

题目:
已知:(x+2)2+|y+1|=0,求5xy2-2xy2-[3xy2-(4xy2-2x2y)]的值.
答案
解:∵(x+2)2+|y+1|=0,
∴x+2=0,y+1=0,
解得x=-2,y=-1,
∵5xy2-2xy2-[3xy2-(4xy2-2x2y)],
=5xy2-2xy2-[3xy2-(4xy2-2x2y)],
=3xy2-3xy2+(4xy2-2x2y),
=4xy2-2x2y,
当x=-2,y=-1时,原式=4xy2-2x2y,
=4×(-2)×(-1)2-2×(-2)2(-1),
=0.
解:∵(x+2)2+|y+1|=0,
∴x+2=0,y+1=0,
解得x=-2,y=-1,
∵5xy2-2xy2-[3xy2-(4xy2-2x2y)],
=5xy2-2xy2-[3xy2-(4xy2-2x2y)],
=3xy2-3xy2+(4xy2-2x2y),
=4xy2-2x2y,
当x=-2,y=-1时,原式=4xy2-2x2y,
=4×(-2)×(-1)2-2×(-2)2(-1),
=0.
考点梳理
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.
根据非负数的性质可得出x和y的值,将原式整理为最简整式后再将x和y的值代入可得出答案.
本题考查整式的化简求值,化简求值是课程标准中所规定的一个基本内容,它涉及对运算的理解以及运算技能的掌握两个方面,也是一个常考的题材.
计算题.
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