數學 : 取值範圍

圖片參考：http://imgcld.yimg.com/8/n/HA06399860/o/701103200043513873392480.jpg

But I don't know why my answer is not the same as your given answer!

2011-03-20 13:37:13 補充：
Which step is wrong? I can't find.

2011-03-20 22:40:22 補充：
解：a^2-bc-8a+7=0---(1)
b^2+c^2+bc-6a+6=0--(2)
(2)-(1)，得：
b^2+c^2+2bc-a^2+2a-1=0
(b+c)^2=(a-1)^2
b+c=±(a-1)-----(3)
(1)×3+(2)，得：
3a^2-3bc-24a+21+b^2+c^2+bc-6a+6=0
3a^2-30a+27+(b-c)^2=0
(b-c)^2=-3(a^2-10a+9)
b-c=±√(-3(a-9)(a-1) )---(4)
因為b和c是實數，所以-3(a-9)(a-1)≥0
(a-9)(a-1)≤0
1≤a≤9

2011-03-20 23:47:07 補充：
First, (3)是沒有用的，sorry for "painting the lily".
Second, I still don't know why my first answer is wrong.
Third, I can only say that the first way is (2)+(1), the second way is (2)-(1).
Fourth, I also want to know why my first answer is wrong.

x^2 + y^2 + z^2 = 0 => x=y=z=0
But x^2 = - (y^2 + z^2) <= 0 will lead to a bigger range for x^2
So this method must be verified with care

2011-03-21 21:56:56 補充：
b+c=±(a-1)-----(3)
b-c=±√(-3(a-9)(a-1) )---(4)
Both these equations are necessary as they explicitly solve b and c in terms
of a, and nothing is missed.

2011-03-21 21:57:31 補充：
for all value in [1,9], b and c are real. Otherwise they are not real, no ambiguity.

實數a，b，c滿足
a^2－bc－8a+7= 0，b^2 + c^2 + bc－6a + 6 = 0，求 a 的取值範圍。
Sol
(b^2+c^2+bc－6a+6)+3(a^2－bc－8a+7)=0
b^2+c^2+3a^2－2bc－30a+27=0
(b-c)^2+3(a^2－10a+9)=0
a^2－10a+9<=0
(a－1)(a－9)<=0
1<=a<=9

a² - bc - 8a + 7 = 0
重寫為
a² - 8a + 7- bc = 0
因為a為實數
所以
8^2-4(7-bc)>0
之後用同一個方法計另一條式

So you are wrong !

2011-03-20 13:44:35 補充：
I donn't know which step is wrong!
But I know b , c have no real roots when a = 13 .

2011-03-20 13:45:05 補充：
So you must be wrong!

2011-03-20 14:18:31 補充：
002乜你講D唔講D?

即係點?

誰能給我完整詳解啊??