證明 2分之1{(x+2)(x-1)-(x-2)(x+1)} =x是恆等式
應該係乘先定係點??
請教恆等式!!!1條
2010-01-17 6:56 am
回答 (1)
2010-01-17 7:17 am
✔ 最佳答案
證明(1/2)[(x+2)(x-1) - (x-2)(x+1)] = xLHS = (1/2)[(x+2)(x-1) - (x-2)(x+1)]
= (1/2)[(x^2 + 2x - x - 2) - (x^2 - 2x + x - 2)]
= (1/2)[(x^2 + x - 2) - (x^2 - x - 2)]
= (1/2)(x^2 + x - 2 - x^2 + x + 2)
= (1/2)(2x)
= x
= RHS
2010-01-16 23:40:45 補充：
002: 你可以從Yahoo寫信給我
收錄日期: 2021-04-23 23:22:40
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100116000051KK01709