F.2 Maths(20points)

1.) 3(a-b)(3r-t) + 9(b-a)
=3(a-b)(r-t)-9(a-b)
=3(a-b)(r-t-9)

2.) (m-n)^2 + n(n-m)
=(m-n)^2-n(m-n)
=(m-n)(m-n-n)
=(m-n)(m-2n)

3.) -ab(rx-s) - ab^2 (s-rx)^2
=ab(s-rx)-ab^2(s-rx)^2
=ab(s-rx)[1-b(s-rx)]
=ab(s-rx)(1-bs+brx)

4.) (5ab-10a) / (15a)
=5a(b-2)/15a
=(b-2)/3

5.) 5x / 6y divide by 15x / 2y
(5x / 6y)/(15x / 2y)
=5x/6y*2y/15x
=5x/15x*2y/6y
=1/3*1/3
=1/9

6.) (2x) / {x(x-1)} plus (5) / (-3x)
(2x) / {x(x-1)} * (5) / (-3x)
=2/(x-1)*5/(-3x)
=10/(x-1)(-3x)
=10/3x(1-x)
[=10/3x-3x^2]

1.)3(a-b)(r-t-9)

2.)(m-n)(m-2n)

3.)ab(s-rx)(1-bs+brx)

4.)(b-2)/3

5.)1/9

6.)[10/3x-3x^2]

I think it is very good for you......

3(a-b)(3r-t-3)
(m-n)(m-2n)
ab(rx-s)(b-1)
(b-2)/3
1/6
1/3x(1-3x)
Wish this can help you

Factorize the following expressions

1. 3(a-b)(3r-t) + 9(b-a)
= 3(a-b)(3r-t) - 9(a-b)
= 3(a-b)(3r-t) - 3[3(a-b)]
= 3(a-b) [3r-t-3]

2. (m-n)^2 + n(n-m)
= (m-n)(m-n) - n(m-n)
= (m-n) (m-n-n)
= (m-n) (m-2n)

3. -ab(rx-s) - ab^2 (s-rx)^2
= ab(s-rx) - ab^2 (s-rx)(s-rx)
= ab(s-rx) [1-b(s-rx)]

Simplify the following expressions.

4. (5ab-10a) / (15a)
= 5a(b-2) / 15a
= (b-2) / 3

5. 5x / 6y divide by 15x / 2y
= 5x / 6y x 2y / 15x
= 1 / 9

6. (2x) / {x(x-1)} plus (5) / (-3x)
= (2x) / [(x) (x-1)] - (5) / (3x)
= [(2x) (3x)] / [(x) (x-1) (3x)] - [(5) (x) (x-1)] / [(3x) (x) (x-1)]
= [6x^2] / [(x) (x-1) (3x)] - [(5x^2 - 5x)] / [(3x) (x) (x-1)]
= [6x^2 - (5x^2 - 5x)] / [(3x) (x) (x-1)]
= [x ^2 + 5x] / [(3x) (x) (x-1)]
= [x (x+5)] / [(3x) (x) (x-1)]
= [x + 5] / [(3x) (x-1)]

2007-09-13 21:57:35 補充：
Factorize 的秘決: x - y = - (y x)Simplify 的秘決: 第4及第5題: 將分子及分母先factorize, 若分子及分母有些項目是一樣, 就可以除消第6題: 通分母, 再用第4及第5題的方法另外, 以下題目可以再簡單化3. ab(s-rx) [1-b(s-rx)] = ab(s-rx)[1-bs-brx]6. [x 5] / [(3x) (x-1)] = [x 5] / [3x^2-3x]

Key to solve 1 - 3:
(a - b) = -(b - a)
1. 3(a - b)(3r - t) + 9(b - a)
= 3(a - b)(3r - t) - 9(a - b)
= 3(a - b) (3r - t - 3)
2. (m - n)2 + n(n - m)
= (m - n)2 - n(m - n)
= (m - n)(m - n - n)
= (m - n)(m - 2n)
3. -ab(rx - s) - ab2 (s - rx)2
= -ab(rx - s) - ab2 (rx - s)2 　　　　　Note: (a - b)2 = [ -(b - a)]2 = (b - a)2
= -ab(rx - s) [1 + ab(rx - s)]
= -ab(rx - s) (1 + abrx - abs)
Key to solve 4 - 6:
Factorize the numerator first. Then cancel out like terms.
4. (5ab - 10a) / 15a
= [5a(b - 2a)] / 15a
= (b - 2a) / 3
5. (5x / 6y) / (15x / 2y)
= (5x)(2y) / [ (6y)(15x) ]
= 1/9
6. 2x / [ x(x - 1) ] + 5 / -3x
= -6x / [ -3x(x - 1) ] + 5(x - 1)/ [ -3x(x - 1) ]
= [5x(x - 1) - 6x ] / [ -3x(x - 1) ]
= x(5x - 11) / [ -3x(x - 1) ]
= (5x - 11) / [ -3(x - 1) ]

第1題: 3(a-b)(3r-t-3)
第2題: (m-n)(m-2n)
第3題: ab(rx-s)(b-1)
第4題: (b-2)/3
第5題: 1/6
第6題: 1/3x(1-3x)

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