2題數學題THX F.1
解下列方程
NO.1
4(2b+5)=13–5(9+b)
NO.2
21-4b=11b–[3(10b–7)+45]
要步驟.
無限THX
回答 (7)
✔ 最佳答案
1.
4(2b+5)=13–5(9+b)
4(2b+5)=13–5(9+b)
8b+20=13-45-5b
8b+5b=13-45-20
13b=-52
b=-4
2.
21-4b=11b–[3(10b–7)+45]
21-4b=11b–[3(10b–7)+45]
21-4b=11b-(30b-21)+4 5
21-4b=11b-30b+21+45
21-21-45=11b-30b+4b
-45=-15b
15b=45
b= 3
參考: me
NO.1
4(2b)+4(5)=13-5(9)+5(b)
8b+20=13-45+5b
8b+20-20=13-45+5b-20
8b= -52+5b
8b-5b=-52+5b-5b
3b=-52
3b/3=-52/3
b=-52/3
NO.2
21-4b=11b-[3(10b–7)+45]
21-4b=11b-[3(10b)–3(7)+45]
21-4b=11b-(30b-21+45)
21-4b=11b-(30b+24)
21-4b=11b-30b-24
21-4b=-19b-24
21-4b+24=-19b-24+24
45-4b+4b=-19b+4b
45=-15b
-15b=45
-15b/-15=45/-15
b=-3
參考: MYSELF
1.
4(2b+5)=13–5(9+b)
8b + 20 = 13 - 45 - 5b
8b + 5b = 13 - 45 - 20
13b = -52
b = -4
2.
21-4b=11b–[3(10b–7)+45]
21 - 4b = 11b - [30b - 21 + 45]
21 - 4b = 11b - 30b +24
-4b - 11b + 30b = 24 - 21
15b = 3
b = 3/15 = 1/5
2007-10-10 16:23:36 補充:
第2題錯左....2.21-4b=11b–[3(10b–7)+45]21 - 4b = 11b - [30b - 21 + 45]21 - 4b = 11b - [30b +24]21 - 4b = 11b - 30b -24-4b - 11b + 30b = -24 - 2115b = -45b = -3
參考: me
1)) 4(2b+5)=13–5(9+b)
8b+20=13-45-5b
8b+5b=13-45-20
13b=-52
b=-4
2)) 21-4b=11b–[3(10b–7)+45]
21-4b=11b-(30b-21)+45
21-4b=11b-30b+21+45
21-21-45=11b-30b+4b
-45=-23b
23b=45
b=45/23
參考: ME
no.1
8b+20=72+8b
8b+8b=20+72
16b=92
no.2
21-4b=11b-(30b-21+45)
21-4b=11b-30b+24
21-4b=-19b+24
-23b=45
1.4(2b+5)=135(9+b)
8b+5=13-45+5b
8b+5=–32+5b
3b=-37
b=-12 1/3
2.21-4b=11b–[3(10b–7)+45]
21-4b=11b-(30b-7+45)
21-4b=-19b-38
59=23b
b=2 13/23
1.
4(2b+5)=13–5(9+b)
8b+20=13-45-5b
8b+5b=13-45-20
13b=-52
b=-4
2.
21-4b=11b–[3(10b–7)+45]
21-4b=11b-(30b-21)+45
21-4b=11b-30b+21+45
21-21-45=11b-30b+4b
-45=-23b
23b=45
b=45/23
參考: my teacher
收錄日期: 2021-04-13 13:49:53
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