更新1:
(Ax+B)(3x-4)≡6x²-23x+20求常數A和B的值
F.2數學題(Ax+2-B(x-3)≡3x+14求常數A和B的值)
回答 (3)
2008-06-08 7:57 pm
✔ 最佳答案
左方=Ax+2-B(x-3)=Ax-Bx+2+3B
=(A-B)x+2+3B
比較左方和右方同類項,可得
A-B=3_____________(1)
2+3B=14___________(2)
從(2)得,
B=4
把B=4代入(1),可得
A-(4)=5
A=9
∴A=9,B=4 //
2008-06-08 12:05:01 補充：
把B=4代入(1),可得
A-(4)=3
A=7
∴A=7,B=4 //
2008-06-08 7:42 pm
Ax+2-B(x-3)≡3x+14
Ax-Bx+3B≡3x+12
(A-B)x+3B≡3x+12
B = 12/3 = 4
A = 3+4 = 7
2008-06-08 12:28:12 補充：
3B≡12
(A-B)x≡3x
Ax-Bx+3B≡3x+12
(A-B)x+3B≡3x+12
B = 12/3 = 4
A = 3+4 = 7
2008-06-08 12:28:12 補充：
3B≡12
(A-B)x≡3x
參考: 自己
2008-06-08 7:42 pm
2 + 3B = 14
3B = 12
B = 4
A - B = 3
Because B = 4
A - 4 = 3
A = 7
You have to expand the left hand side and group the like terms and compare
them with the right hand side, NEVER transpose terms in both sides.
Remember, this is an idnetity, not an equation.
3B = 12
B = 4
A - B = 3
Because B = 4
A - 4 = 3
A = 7
You have to expand the left hand side and group the like terms and compare
them with the right hand side, NEVER transpose terms in both sides.
Remember, this is an idnetity, not an equation.
參考: myself
收錄日期: 2021-04-23 20:33:37
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