f.2maths of formulae

2008-01-06 4:32 am
factorize this expression.
no.1. 3(a-b)(3r-t)+9(b-a)(2t-5s)

no.2 . (s+1)t-1-s

no.3. (2x+1)(x-y)-(3x+1)(x-y)

no.4. (m-n)^2+n(n-m)

no.5. -ab(rx-s)-ab^2(s-rx)^2
(請有計算過程和答案)

回答 (2)

2008-01-06 5:06 am
✔ 最佳答案
1)3(a-b)(3r-t)+9(b-a)(2t-5s)
=3(a-b)(3r-t)-9(a-b)(2t-5s)
=3(a-b)[(3r-t)-3(2t-5s)]
=3(a-b)(3r-t-6t+15s)
=3(a-b)(3r-7t+15s)

2)(s+1)t-1-s
=(s+1)t-(s+1)
=(s+1)(t-1)

3)(2x+1)(x-y)-(3x+1)(x-y)
=(x-y)[(2x+1)-(3x+1)]
=(x-y)(2x+1-3x-1)
=-x(x-y)

4)(m-n)^2+n(n-m)
=(m-n)^2-n(m-n)
=(m-n)[(m-n)-n]
=(m-n)(m-2n)

5)-ab(rx-s)-ab^2(s-rx)^2
=ab(s-rx)-ab^2(s-rx)^2
=(s-rx)[ab-ab^2(s-rx)]
=ab(s-rx)[1-b(s-rx)]
=ab(s-rx)(bs-rx+1)
2008-01-06 4:43 am
1.
3(a-b)(3r-t)+9(b-a)(2t-5s)
= (3a-3b)(3r-t)+(9b-9a)(2t-5s)
= 3r(3a-3b)-t(3a-3b)+9b(2t-5s)-9a(2t-5s)
= 9ar-9br-3at+3bt+18bt-45bs-18at+45as
= 9ar-21at+45as-9br+21bt-45bs

睇到眼花, 等我透下氣繼續= =

2008-01-05 21:01:13 補充:
2 . (s 1)t-1-s= ts t-1-s= t ts-s-13. (2x 1)(x-y)-(3x 1)(x-y)= x(2x 1)-y(2x 1)-x(3x 1) y(3x 1)= 2x^2 x-2xy-y-3x^2-x 3xy y= -1x^2 xy4. (m-n)^2 n(n-m)= m^2-n^2 n^2-mn= m^2-mn5. -ab(rx-s)-ab^2(s-rx)^2= -abrx abs-[ab^2s ab^2rx]^2= -abrx abs-ab^4.... 因析分解未學, sorry


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