急!超易!送嚸啊!中一MATHS题!20嚸啊!!!

2007-09-28 2:25 am
要有步爪:
7k-(9k-16)=14-(19-k)

3[4(2.5-q)+1]=-2

2[p-2(p+1)-1=8p+25

回答 (3)

2007-09-28 2:34 am
✔ 最佳答案
7k - (9k - 16) = 14 - (19 - k)
7k - 9k + 16 = 14 - 19 + k
3k = 21
k = 7
-------------------------------------

3[4(2.5 - q) + 1] = -2
3[10 - 4q + 1] = -2
3[11 - 4q] = -2
33 - 12q = -2
12q = 35
q = 35/12
-------------------------------------

2[p - 2(p + 1) - 1] = 8p + 25
2[p - 2p - 2 - 1] = 8p + 25
2[-p - 3] = 8p + 25
-2p - 6 = 8p + 25
10p = -31
p = -31/10
2007-09-28 2:46 am
7k-(9k-16)=14-(19-k)
7k-9k+16=14-19+k
-2k= -5-16+k
5+16=2k+k
21=3k
7=k
2007-09-28 2:35 am
1/
7k-(9k-16)=14-(19-k)
7k - 9k +1 =14 -19 +k
3k = -5 -1
k = -6/3
k= -2

2/
3[4( 2.5 -q )+1]= -2
12( 2.5 -q )+3= -2
30 - 12q = -5
12q = 35
q = 35/12

3/
2[p -2(p+1) -1]=8p+25
2[ p -2p -2 -1 ] = 8p +25
2( -p -3 ) = 8p +25
-2p -6 = 8p +25
10p = -31
p = -31/10
參考: myself


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