1.Ax-3≡8x+B,求A,B值
2.(x+3)(x-2)≡Ax^2+Bx+C,求A,B,C值
更新1:
打錯題目.... 第1題應該是 3Ax+7≡6x-B,求A,B值 SORRY
恆等式一問
回答 (6)
2006-11-09 3:44 am
✔ 最佳答案
恆等式 (identity)更正第1題前的回答：
1.若Ax﹣3≡8x+B,求A,B值
左方RHS
=Ax﹣3
右方LHS
=8x+B
∵Ax﹣3≡8x+B
∴A = 8，B = ﹣3
==============================
更正第1題後的回答：
1.若3Ax+7≡6x﹣B,求A,B值
左方RHS
=3Ax+7
右方LHS
=6x﹣B
∵3Ax+7≡6x﹣B
∴3A = 6
A = 6/3
A = 2
∴﹣B = 7
B = ﹣7
2.若(x+3)(x﹣2)≡Ax^2+Bx+C,求A,B,C值
左方RHS
=(x+3)(x﹣2)
=x2﹣2x+3x﹣6
=x2 + x﹣6
右方LHS
=Ax^2+Bx+C
∵(x+3)(x﹣2)≡Ax^2+Bx+C
∴A = 1 ←(1)x2 = x2，所以A是1
∴B = 1 ←(1)x = x，所以B是1
∴C = ﹣6
參考: 中二級數學書
2006-11-09 3:20 am
第1題應該是
-7=B
第2題應該是
A=1
B=1
C=-6
-7=B
第2題應該是
A=1
B=1
C=-6
2006-11-09 2:48 am
Ax-3≡8x+B
左方=Ax-3
右方=8x+B
A=8 B=-3
左方=Ax-3
右方=8x+B
A=8 B=-3
2006-11-09 2:45 am
1.
A = 8
B = -3
2.
A = 1
B = 1
C = -5
2006-11-08 18:46:48 補充：
打錯左 sor.....2.C = -6
A = 8
B = -3
2.
A = 1
B = 1
C = -5
2006-11-08 18:46:48 補充：
打錯左 sor.....2.C = -6
2006-11-09 2:45 am
1. Ax - 3 = 8x + B
i.e. A = 8 and B = -3
2. (x + 3)(x - 2) = Ax^2 + Bx + C
x^2 + x - 6 = Ax^2 +Bx + C
i.e. A = 1, B = 1 and C = -6
i.e. A = 8 and B = -3
2. (x + 3)(x - 2) = Ax^2 + Bx + C
x^2 + x - 6 = Ax^2 +Bx + C
i.e. A = 1, B = 1 and C = -6
2006-11-09 2:44 am
1.Ax-3≡8x+B,求A,B值
A=8,B= -3
2.(x+3)(x-2)≡Ax²+Bx+C,求A,B,C值
LHS=x²+x-6
so A=1,B=1,C= -6
A=8,B= -3
2.(x+3)(x-2)≡Ax²+Bx+C,求A,B,C值
LHS=x²+x-6
so A=1,B=1,C= -6
收錄日期: 2021-04-23 15:58:29
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