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2007-09-22 7:24 pm
t^2-20h+100


(a+b)^2 - 2(a+b) - 15


2u^2 +6h-140

回答 (3)

2007-09-22 7:33 pm
✔ 最佳答案
1)
Assume that you type the question wrong...
t^2 - 20t + 100
= (t - 10)(t - 10)
= (t-10)^2

2)
(a+b)^2 - 2(a+b) - 15
You can let y = a+b
The whole equation becomes y^2 - 2y - 15
= (y - 5)(y + 3)
Put y=a+b back to the equation
= (a+b-5)(a+b+3)

3)
Assume that you type the question wrong again...
2u^2 + 6u - 140
= 2(u^2 + 3u - 70)
= 2(u+10)(u-7)
參考: Me
2007-09-22 8:21 pm
1)t^2-20t+100
=t^2-(2)(10)+10^2
=(t-10)^2

2)(a+b)^2 - 2(a+b) - 15
=[(a+b)-5][(a+b)+3]
=(a+b-5)(a+b+3)

3)2u^2 +6u-140
=2(u^2+3u-70)
=2(u-7)(u+10)
2007-09-22 7:36 pm
t^2-20h+100
=(t-10)^2

(a+b)^2 - 2(a+b) - 15
=[(a+b)-5][(a+b)+3]

2u^2 +6h-140
=(2u-14)(u+10)


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