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试题原文 |
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解:(1)原式=﹣32+16+1=﹣15; (2)原式= = =﹣24b2c5; (3)原式=﹣1+4+1=4; (4)原式=[x2+4xy+4y2﹣(3x2﹣xy+3xy﹣y2)﹣5y2]÷2x =[x2+4xy+4y2﹣3x2﹣2xy+y2﹣5y2]÷2x =[﹣2x2+2xy]÷2x =﹣x+y; (5)原式=4x2﹣(y﹣1)2 =4x2﹣(y2﹣2y+1) =4x2﹣y2+2y﹣1; (6)原式=[(x﹣y)(x2+y2)(x+y)]2 =[(x2﹣y2)(x2+y2)]2 =[x4﹣y4]2 =x8﹣2x4y4+y8; (7)原式=8x3(﹣2y3)÷(16xy2)=﹣x2y; (8)原式=(y﹣2x)2﹣(y﹣2x)(y+2x)+(x﹣2y)(y﹣2x) =(y﹣2x)(y﹣2x﹣y﹣2x+x﹣2y) =(y﹣2x)(﹣2y﹣3x) =﹣2y2﹣3xy+2xy+6x2 =﹣2y2﹣xy+6x2; (9)原式=[4x4y2xy2﹣6x3x3y6]÷(﹣2x4y4) =[4x5y4﹣6x6y6]÷(﹣2x4y4) =﹣2x+3x2y2. |
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“计算:(1).(2).(3).(4)[(x+2y)2﹣(x+y)(3x﹣y)﹣5y2]÷2x.(5)(2x﹣y+1)..”的主要目的是检查您对于考点“初中整式的加减乘除混合运算”相关知识的理解。有关该知识点的概要说明可查看:“初中整式的加减乘除混合运算”。