１ａ．以證明下列各式是恆等式。
＊＊要列式！＊＊
題目在下面：
2(s+7)+3(s-5)=s+4(s-2)+7
ｂ.(2x+3)(x-2)=2x(x+6)-13x-6
我想問一些ｍａｔｈｓ？
2006-12-27 7:12 am
回答 (4)
2006-12-27 7:19 am
✔ 最佳答案
2(s+7)+3(s-5)=s+4(s-2)+7L.H.S = 2(s+7)+3(s-5)
= 2s + 14 + 3s - 15
= 2s + 3s + 14 - 15
= 5s - 1
R.H.S = s+4(s-2)+7
= s + 4s - 8 + 7
= 5s - 1
(2x+3)(x-2)=2x(x+6)-13x-6
L.H.S = (2x+3)(x-2)
= 2x^2 - 4x + 3x - 6
= 2x^2 -x - 6
R.H.S = 2x(x+6)-13x-6
= 2x ^ 2 + 12x - 13x - 6
= 2x^2 - x - 6
2006-12-27 7:53 am
2(s+7)+3(s-5)=s+4(s-2)+7
L.H.S = 2(s+7)+3(s-5)
= 2s + 14 + 3s - 15
= 2s + 3s + 14 - 15
= 5s - 1
R.H.S = s+4(s-2)+7
= s + 4s - 8 + 7
= 5s - 1
(2x+3)(x-2)=2x(x+6)-13x-6
L.H.S = (2x+3)(x-2)
= 2x^2 - 4x + 3x - 6
= 2x^2 -x - 6
R.H.S = 2x(x+6)-13x-6
= 2x ^ 2 + 12x - 13x - 6
= 2x^2 - x - 6
L.H.S = 2(s+7)+3(s-5)
= 2s + 14 + 3s - 15
= 2s + 3s + 14 - 15
= 5s - 1
R.H.S = s+4(s-2)+7
= s + 4s - 8 + 7
= 5s - 1
(2x+3)(x-2)=2x(x+6)-13x-6
L.H.S = (2x+3)(x-2)
= 2x^2 - 4x + 3x - 6
= 2x^2 -x - 6
R.H.S = 2x(x+6)-13x-6
= 2x ^ 2 + 12x - 13x - 6
= 2x^2 - x - 6
2006-12-27 7:20 am
2(s+7)+3(s-5)=s+4(s-2)+7
2s+14+3s-15=s+4s-8+7
5s-1=5s-1
2s+14+3s-15=s+4s-8+7
5s-1=5s-1
2006-12-27 7:20 am
No. 1
L.H.S : 2[s+7]+3[s-5]=2s+14+3s-15
= 5s -1
R.H.S: s+4(s-2)+7 = s+4s-6+7
=5s-1
=L.H.S
No.2
L.H.S: .(2x+3)(x-2)=2x^2 +3x -4x-6
=2x^2 -x -6
R.H.S: 2x(x+6)-13x-6=2x^2 +12x-13x-6
=2x^2-x-6
=L.H.S
L.H.S : 2[s+7]+3[s-5]=2s+14+3s-15
= 5s -1
R.H.S: s+4(s-2)+7 = s+4s-6+7
=5s-1
=L.H.S
No.2
L.H.S: .(2x+3)(x-2)=2x^2 +3x -4x-6
=2x^2 -x -6
R.H.S: 2x(x+6)-13x-6=2x^2 +12x-13x-6
=2x^2-x-6
=L.H.S
收錄日期: 2021-04-12 22:34:29
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061226000051KK05098