发布人:繁体字网(www.fantiz5.com) 发布时间:2015-03-23 07:30:00
试题原文 |
|
(1)(4xy-x2-y2)-(x2-y2+6xy) =4xy-x2-y2-x2+y2-6xy =(-1-1)x2+(-1+1)y2+(4-6)xy =-2x2-2xy; (2)x(x-1)+(2x+5)(2x-5) =x2-x+4x2-25 =5x2-x-25; (3)[(x2+y2)-(x-y)2+2y(x-y)]÷2y =[x2+y2-(x2-2xy+y2)+2xy-2y2]÷2y =(x2+y2-x2+2xy-y2+2xy-2y2)÷2y =(-2y2+4xy)÷2y =-y+2x…(2分) 由于2x=y+15,则-y+2x=15,代入原式=15. |
经过对同学们试题原文答题和答案批改分析后,可以看出该题目“(1)计算:(4xy-x2-y2)-(x2-y2+6xy)(2)计算:x(x-1)+(2x+5)(2x-5)(3..”的主要目的是检查您对于考点“初中整式的加减”相关知识的理解。有关该知识点的概要说明可查看:“初中整式的加减”。