✔ 最佳答案
證明以下各恆等式 a)(a-b)(c-d)+(a-d)(b -c)<恆等符號>(a-c)(b-d)
(a-b)(c-d)+(a-d)(b -c)
= ac - ad - bc + bd + ab - ac - bd + cd
= ab - ad + cd - bc
= a(b-d) + c(d-b)
= (a-c)(b-d)
左右相等
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2)以代入法解下列方程組(二元一次) a)16w = 5-4s
3s-8=8w+7
16w = 5-4s --------- (1)
3s=8w+15
s = (8w+15)/3
代入 (1), 16w = 5-4(8w+15)/3
48w = 15-4(8w+15)
48w = 15-32w-60
80w = -45
w = -9/16
s = (8w+15)/3 = 21/6
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3)以消去法解下列方程組(二元一次) a)6x- y-4」3 =10 b)5x+6y =1
y+ 3x-7」4=1 4」3(x) - 2y= 8」3
c)5」m + 6」n=1
4」3m-2」n=8」3
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4)三個正數的比例為3:4:5,若最大和最小的數相差10,求這三個數
設三個數為 3x, 4x, 5x
5x - 3x = 10
x = 5
三個數為 15, 20, 25
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5)把以下十六進數轉為二進數:
a)18A (16) = 110001010 (2)
b)ABC (16) = 101010111100 (2)