國二數學代數問題

2013-12-27 7:21 am
1.已知√(x-2)=6-【9/√(x-2)】,則x=?
2.(1/2)<a<(2/3),,-(1/5)<b<(1/3),且x=(a+b)^2+1,y=2a+2b,則√(x+y) + √(x-y)=?

會一題也可以回答

回答 (4)

2013-12-27 7:51 pm
✔ 最佳答案
1.已知√(x-2)=6-[9/√(x-2)],則x=?
Sol
Set √(x-2)=y
y=6-9/y
y^2=6y-9
y^2-6y+9=0
(y-3)^2=0
y=3(重根)
√(x-2)=3
x-2=9
x=11

2.1/2<a<2/3,-1/5<b<1/3,且x=(a+b)^2+1,y=2a+2b,則√(x+y)+√(x-y)=?
Sol
1/2<a<2/3,-1/5<b<1/3
1/2-1/5<a+b<2/3+1/3
3/10<a+b<1
13/10<a+b+1<2
-7/10<a+b-1<0
x+y
=(a+b)^2+1+2a+2b
=(a+b)^2+2(a+b)+1^2
=(a+b+1)^2
x-y
=(a+b)^2+1-2a-2b
=(a+b)^2-2(a+b)+1^2
=(a+b-1)^2
√(x+y)+√(x-y)
=|a+b+1|+|a+b-1|
=(a+b+1)-(a+b-1)
=2


2013-12-27 5:23 pm
1.令Y=(x-2)^(1/2),則題目寫成Y=6-(9/Y)
,左右同乘Y得到Y^2=6Y-9,移項之後得到一個完全平方式:(Y-3)^2=0,解出Y=3
所以(x-2)^(1/2)=3,平方之後x-2=9,x=11

2013-12-27 09:43:59 補充:
2.(3/10)<(11/6),(9/100)<(a+b)^2<(121/36),(109/100)<(a+b)^2+1=x<(157/36)
(3/5)<2(a+b)=y<(11/3),(13/10)<(x+y)^(1/2)<(17/6)
,(773^(1/2)/10*3^(1/2))<(x-y)^(1/2)<(677^(1/2)/6*5^(1/2))
,(773^(1/2)/10*3^(1/2))+(13/10)<(x-y)^(1/2)+(x+y)^(1/2)<(677^(1/2)/6*5^(1/2))+(17/6)
2013-12-27 9:09 am
1.x=11...........

2013-12-27 01:14:19 補充:
2.

2(a+b)不等於2ab.......
2013-12-27 7:45 am
1.令√(x-2)=y
y=6-(9/y) y-6+(9/y)=0 y^2-6y+9=0 (y-3)^2=0 y=3
y=3 √(x-2)=3 x=5

2.a+b>0 x=(a+b)^2+1,y=2(a+b)
√(x+y) =√((a+b)^2+2(a+b)+1) =a+b √(x-y)==√((a+b)^2-2(a+b)+1)=a-b
√(x+y) + √(x-y)=(a+b)+(a-b)=2a


收錄日期: 2021-04-30 18:18:37
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