F1數學方程式

2007-12-18 7:01 pm
(3x+4)(4x+7)/2 - (3x+13)(6x+3)/3 =1

回答 (2)

2007-12-18 7:16 pm
✔ 最佳答案
(3x+4)(4x+7)/2 - (3x+13)(6x+3)/3 =1
左右各乘以6 (天秤法)
3 (3x+4)(4x+7) - 2(3x+13)(6x+3) = 6
分別將兩組括號展開
3 (12x^2 + 37 x+ 28) - 2(18x^2+87x+39) = 6
數字乘入括號
(36x^2 + 111 x+ 84) - (36x^2+174x+78) = 6
拆括號
36x^2 + 111 x+ 84 - 36x^2 - 174x - 78 = 6
84x + 6 = 6
84 x = 0
x = 0
2007-12-18 7:31 pm
跟據benny_in_india的回答

直致以下都正確

(3x+4)(4x+7)/2 - (3x+13)(6x+3)/3 =1
左右各乘以6 (天秤法)
3 (3x+4)(4x+7) - 2(3x+13)(6x+3) = 6
分別將兩組括號展開
3 (12x^2 + 37 x+ 28) - 2(18x^2+87x+39) = 6
數字乘入括號
(36x^2 + 111 x+ 84) - (36x^2+174x+78) = 6
拆括號
36x^2 + 111 x+ 84 - 36x^2 - 174x - 78 = 6

但之後應該是如下

-63x+6 = 6
63x = 12
x = 12/63
x = 0.190476.......
參考: me


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