F.2數學

1.化簡(8m-7m[二次]+3)-(m[二次]-2)+(1+8m[二次]-9m)
(8m-7m^2+3)-(m^2-2)+(1+8m^2-9m)
= 8m - 7m^2 + 3 - m^2 + 2 + 1 + 8m^2 - 9m
= 6 - m

2.展開(x[二次]-2x-3)(x+5),答案依變數的降冪排列
(x^2-2x-3)(x+5)
= (x^2-2x-3)x + (x^2-2x-3)(5)
= x^3-2x^2-3x + 5x^2-10x-15
= x^3 + 3x^2 - 13x - 15

3.志強騎單車走完280m需時25s,志明騎單車走完216m需時18s,誰較快,快多少?
志強車速 = 280/25 = 11.2 m/s
志明車速 = 216/18 = 12 m/s
志明快, 快左 0.8 m/s 即每秒快 0.8 米

4.將1根108cm的銅線屈成一個三角形,3條邊長度的比是3:4:5,求最長一邊的長度
設邊長為 3x, 4x, 5x
3x + 4x + 5x = 108
12x = 108
x = 9
最長一邊的長度 = 5x = 45 cm

2007-08-26 00:14:59 補充：
慢左,不過我答得齊 d,同埋答得清楚 d多謝支持 .............

1. -m+6

2. x^3 +3x^2 - 13x - 15

3. 志強的速度= 11.2 m/s
志明的速度= 12 m/s
志明較快

4. 45 cm

2007-08-26 01:01:22 補充：
最簡潔明確

1. (8m-7m^2+3)-(m^2-2)+(1+8m^2-9m)
=8m-7m^2+3-m^2+2+1+8m^2-9m
=6-2m

2. (x^2-2x-3)(x+5)
=x^3-2x^2-3x+5x^2-10x-15
=x^3+3x^2-13x-15

3. speed of chi keung: 280m/25s = 11.2m/s
speed of chi ming: 216m/18s = 12m/s
hence, chi ming is faster.

4. the longest length of the triangle will be
108 * 5/(3+4+5) = 108 * 5/12 = 45.

2007-08-26 00:11:13 補充：
sorry! answer of 1 is wrong.it should be=8m-7m^2 3-m^2 2 1 8m^2-9m=(-7-1 8)m^2 (8-9)m (3 2 1)=6-m

1. (8m-7m[二次]+3)-(m[二次]-2)+(1+8m[二次]-9m)
= 8m-7m^2+3 - m^2 + 2 +1+8m^2-9m
=m^2(-7-1+8) + m(8-9) + (3+2+1)
=-m+6

2. (x[二次]-2x-3)(x+5)
=x^2(x+5) - 2x(x+5) -3(x+5)
=x^3 + 5x^2 -2x^2 - 10x -3x-15
=x^3 +3x^2 - 13x - 15

3. 志強的速度 = 280/25 = 11.2 m/s
志明的速度 = 216/18 = 12 m/s
所以志明較快

4. 3條邊長分別是 108x3/(3+4+5), 108x4/(3+4+5) 及 108x5/(3+4+5)
= 27, 36 及 45
所以最長的長度是45 cm

1.(8m-7m[二次]+3)-(m[二次]-2)+(1+8m[二次]-9m)
=8m-7m[二次]+3-m[二次]+2+1+8m[二次]-9m
=-m+6
2.(x[二次]-2x-3)(x+5)
=x[三次]-6x[二次]-3x+5x[二次]-10x-15
=x[三次]-x[二次]-13x-15
3.志強:280/25=11.2s
志明:216/18=12s
4.108/(3+4+5)=9
9*5=45