普通恒等式問題

2015-06-29 12:43 am
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

回答 (2)

2015-06-29 1:10 am
✔ 最佳答案
first of all, re-arrange the variables in correct form:
-2(10s+3t)(3t-10s)
= -2[(3t+10s)(3t-10s)]
= -2[(3t)^2 - (10s)^s]
= -2[9t^2 - 100s^2]
= 200s^2 - 18t^2

can't see what's no. 2

for no.3:
3(-7h-2k)(2k-7h)
= 3{[(-7h)-2k][(-7h)+2k]}
= 3[(-7h)^2 - (2k)^2]
= 3[49h^2 - 4k^2]
= 147h^2 - 12k^2
2015-06-29 1:20 am
簡單來說,回答發問同學的問題,錯誤在於寫了
(3t - 10s) 為 (10s - 3t)。

事實上
(3t - 10s) ≠ (10s - 3t)

正確的是
(3t - 10s) = -(10s - 3t) 〔有一個負號〕


收錄日期: 2021-04-30 01:35:05
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