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若 x + z = 0, -y + z = 1 及 3x +
若
x + z = 0
-y + z = 1
3x + y + 2z = -1
有解 (x, y, z) 滿足 (x - p)^2 + y^2 + z^2 = 1,
求 p 值的範圍。
THX!!!
1 Answer
- 匿名Lv 79 years agoFavorite Answer
| 1 0 1 0|
| 0 -1 1 1| ->
| 3 1 2 -1|
| 1 0 1 0|
| 0 -1 1 1| ->
| 3 0 3 0|
| 1 0 1 0|
| 0 -1 1 1|
| 0 0 0 0|
Put z = t,
x + t = 0 => x = -t
-y + t = 1 => y = t-1
{x,y,z} = {-t,t-1,t}, for t is any real number
(-t-p)^2 + (t-1)^2 + t^2 = 1
t^2 + 2tp + p^2 + t^2 - 2t + 1 + t^2 = 1
3t^2 + t(2p-2) + p^2 = 0
For t has >= 1 solution,
△ >= 0
(2p-2)^2 - 4(p^2)(3) >= 0
4p^2 - 8p + 4 - 12p^2 >= 0
-8p^2 - 8p + 4 >= 0
2p^2 + 2p - 1 <= 0
(-2-√12)/4 <= p <= (-2+√12)/4
(-1-√3)/2 <= p <= (-1+√3)/2
Source(s): Hope the solution can help you^^”