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试题原文 |
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解:(1)(﹣)0+52+(﹣)﹣2+(﹣2)3 =1+25+4+(﹣8) =30﹣8 =22; (2)x2x4+(﹣x3)2﹣x7÷x=x6+x6﹣x6=x6; (3)(xy)2(﹣12x2y2)÷(﹣x3y) =x2y2(﹣12x2y2)(﹣) =4xy3; (4)5x(2x2﹣3x+4) =10x3﹣15x2+20x; (5)(3x+5)(3﹣2x) =9x﹣6x2+15﹣10x =﹣6x2﹣x+15; (6)(﹣2a﹣1)2 =(2a+1)2 =4a2+4a+1; (7)(x+2)(y+3)﹣(x+1)(y﹣2) =xy+3x+2y+6﹣(xy﹣2x+y﹣2) =xy+3x+2y+6﹣xy+2x﹣y+2 =5x+y+8; (8)(﹣2p﹣q+1)(﹣q+2p﹣1) =[﹣q﹣(2p﹣1)][﹣q+(2p﹣1)] =(﹣q)2﹣(2p﹣1)2=q2﹣(4p2﹣4p+1) =q2﹣4p2+4p﹣1; (9)(x﹣y)2﹣(x+y)(x﹣y) =(x2﹣2xy+y2)﹣(x2﹣y2) =x2﹣2xy+y2﹣x2+y2 =2y2﹣2xy; (10)[(x+y)2﹣(x﹣y)2]÷(2xy) =[(x2+2xy+y2)﹣(x2﹣2xy+y2)]÷2xy =4xy÷2xy =2. |
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)x2x4+(﹣x3)2﹣x7÷x(3)(4)5x(2x2﹣3x+4)(5)(3x+5)(3﹣2x)..”的主要目的是检查您对于考点“初中零指数幂(负指数幂和指数为1)”相关知识的理解。有关该知识点的概要说明可查看:“初中零指数幂(负指数幂和指数为1)”。