我想問中一數學!

2007-12-31 7:39 pm
我想問既就是以下這幾條:
q1 (2a+1)(a-3)(3a-1)
q2 15-(r-6)=6-(3r+5)
q3 兩個連續整數之和是95。求該兩個數。

回答 (6)

2007-12-31 8:03 pm
✔ 最佳答案
q2.15-(r-6)=6-(3r+5)
15-r-6=6-3r+5
15-6-6-5=3r+r
-2=4r
r= -0.5

q3. 設最小數是y
y+y+1=95
2y=94
y=47

2007-12-31 12:08:43 補充:
q2.15-(r-6)=6-(3r+5)15-r+6=6-3r-515+6-6+5= -3r+r20=-2rr= -10
2007-12-31 8:05 pm
q1 (2a+1)(a-3)(3a-1)
= a(2a+1)-3(2a+1)(3a-1)
= (2a^2+a-6a-3) (3a-1)
= 3a(2a^2-5a-3)-(2a^2-5a-3)
= 6a^3-15a^2-9a-2a^2+5a+3
= 6a^3-17a^2-4a+3

q2 15-(r-6)=6-(3r+5)
15-r+6=6-3r-5
20=-2r
-10=r

q3 兩個連續整數之和是95。求該兩個數。
設:最小的數是x,另一個的數是(x+1),
得:x+(x+1)=95
2x+1=95
2x=94
x=47
參考: myself
2007-12-31 8:04 pm
係唔係要簡化or擴大!你打清楚!如果唔係好難題你wo!
2007-12-31 8:04 pm
q1)2a+6a+6a-2a+a-3+3a-1
=20a-4
q2)15-r+6=6-3r-5
9+6=-4r-5
20=4r
r =5
2007-12-31 7:56 pm
Q3:n+(n+1) =95
參考: me
2007-12-31 7:53 pm
q1同q2要化簡?不過我唔識-.-
q3好easy姐,先將95-1,之後除2,得出47,47+47=94,47+48=95!!!!!

2007-12-31 12:01:28 補充:
15-(r-6)=6-(3r 5) 15-r 6=6-3r-5 21-r=1-3r 21=1-2r 21-21=1-2r-21 2r=20 r=10

2007-12-31 12:11:33 補充:
r係-10至岩-.-,2r=-20


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