用恒等式(x+y)(x-y)≡x^2-y^2 計
1.
-2(10s+3t)(3t-10s)
2.
3(-7h-2k)(2k-7h)
我計的(錯在那)
-2(10s+3t)(3t-10s)
= -2(10s+3t)(10s-3t)
= -2[(10s)^2-3t^2]
= -2(100s^2-9t^2)
= -200s^2-18t^2
收錄日期: 2021-04-30 01:35:05
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20150628000051KK00071