F2 maths (10 points)

1.) ab+3a+b+3
=b(a+1)+3(a+1)
=(a+1)(b+3)

2.) cd-c+d-1
=d(c+1)-(c+1)
=(c+1)(d-1)

3.) hk-h-k+1
=h(k-1)-(k-1)
=(k-1)(h-1)

4.) 1-p–q+pq
=p(q-1)-(q-1)
=(q-1)(p-1)

5.) 1+x+y+xy
=x(y+1)+(y+1)
=(y+1)(x+1)

6.) bx–by+2y-2x
=b(x-y)-2(x-y)
=(x-y)(b-2)

7.) mn-np+4m–4p
=n(m-p)+4(m-p)
=(m-p)(n+4)

1) 首先，你要先把共有的整理。在這題，你可先整理"a, "b"或"3" 都可以：
若先整理"a"，則得出: a(b+3)+b+3 ,
而剛好b+3 可以再整理得出：(b+3)(a+1)

若先整理"b"，則得出: b(a+1)+3a+3 ,
而剛好"3" 可以再整理得出：b(a+1)+3(a+1)
而剛好a+1" 可以再整理得出：(a+1)(b+3)

若先整理"3"，則得出: 3(a+1)+ab+b ,
而剛好"b" 可以再整理得出：3(a+1)+b(a+1)
而剛好a+1" 可以再整理得出：(a+1)(b+3)

你可以看出三種做法的答案都是一樣的，但第一種就簡單一點，多做你就可以掌握，努力！


2) 按(1)的方法先抽"c"來整理，就得出：c(d-1)+d-1
而剛好d-1 可以再整理得出：(d-1)(c+1)

跟著的幾題不再多說，直接做啦！

3) h(k-1)-k+1 => (k-1)(h-1) （注：小心正負號）

4) p(q-1)-q+1 => (q-1)(p-1) （注：小心正負號）

5) x(y+1)+y+1 => (y+1)(x+1)

6) b(x-y)+2(y-x) => (x-y)(b-2) （注：小心2(y-x)變為 -2(x-y)的過程）

7) n(m-p)+4(m-p) => (m-p)(n+4)

1.ab + 3a + b + 3
= a( b +3 ) +( b +3 )
= ( a +1 )( b +3 )

2.cd – c + d – 1
= c ( d -1 ) +( d -1 )
= ( c +1 )( d -1 )

3.hk - h - k +1
= h( k -1 ) -( k -1 )
= ( h -1 )( k -1 )

4.1- p – q + pq
= ( 1 -p ) -q( 1 -p )
= ( 1 -p )( 1 -q )

5.1 + x + y + xy
= ( 1 +x ) +y( 1 +x )
=( 1 +x )( 1 +y )

6.bx – by + 2y - 2x
= b( x -y ) -2( x -y )
= ( b -2 )( x +y )

7.mn - np + 4m – 4p
= n( m -p ) +4( m -p )
= ( n +4 )( m -p )

2007-10-07 22:44:42 補充：
第6條打錯符號, 應該係:6.bx – by 2y - 2x= b( x -y ) -2( x -y )= ( b -2 )( x -y )

1) 3(a+1)+b(a+1)=(3+b)(a+1)
2) c(d-1)+(d-1)=(c+1)(d-1)
3) h(k-1)-(k-1)=(h-1)(k-1)
4) (1-p)-q(1-p)=(1-q)(1-p)
5) (1+x)+y(1+x)=(1+y)(1+x)
6) b(x-y)-2(x-y)=(b-2)(x-y)
7) n(m-p)-4(m-p)=(n-4)(m-p)

1. b(a+1)+3(a+1)=(b+3)(a+1)
2. c(d-1)+(d-1)=(c+1)(d-1)
3. h(k-1)-(k-1)=(h-1)(k-1)
4. (1-p)-q(1-p)=(1-q)(1-p)
5. (1+x)+y(1+x)=(1+y)(1+x)
6. b(x-y)-2(x-y)=(b-2)(x-y)
7. n(m-p)+4(m-p)=(n+4)(m-p)

1.(a+1)(3+b)
2.(c+1)(d-1)
3.(p-1)(q-1)
4.(k-1)(h-1)
5.(x+1)(y+1)
6.(x-y)(b-2)
7.(m-p)(n+4)

2007-10-07 22:41:13 補充：
the answer of no.3 is (k-1)(h-1)the answer of no.4 is (p-1)(q-1)