我想問4((x-0.2)^2=(2x-0.3)^2點計
回答 (3)
2007-09-14 4:20 am
✔ 最佳答案
1. 折括號.4 (x^2 - 2(x)(0.2) +0.2^2) = (2x)^2 - 2(2x)(0.3) + (0.3)^2
4x^2 - 1.6x + 0.04 = 4x^2 -1.2x +0.09
x平方項 抵消. , 全式 X 100 唔要小數點.
-160x + 4 = -120x + 9
x = -0.125
2007-09-24 4:04 am
十分詳細，而且很清楚。
2007-09-14 4:21 am
4(x-0.2)^2=(2x-0.3)^2
4(x^2-0.4x+0.04)=4x^2-1.2x+0.09
4x^2-1.6x+0.16=4x^2-1.2x+0.09
4x^2-4x^2-1.6x+1.2x=0.09-0.16
-0.4x=-0.07
x=0.175
4(x^2-0.4x+0.04)=4x^2-1.2x+0.09
4x^2-1.6x+0.16=4x^2-1.2x+0.09
4x^2-4x^2-1.6x+1.2x=0.09-0.16
-0.4x=-0.07
x=0.175
參考: me
收錄日期: 2021-04-25 21:55:48
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https://hk.answers.yahoo.com/question/index?qid=20070913000051KK03298