X^2 + 16 = 5X + Y .....(1)
K + Y = 5X .......(2)
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回答 (3)
2007-01-09 2:36 am
✔ 最佳答案
Differentiate (1), we have2x=5+(dy/dx)
dy/dx=2x-5 ......(3)
Differentiate (2), we have
0+dy/dx=5
dy/dx=5...........(4)
From (3) and (4),
2x-5=5
x=5.................(5)
sub. (5) into (1)
y=16...............(6)
sub (5) and (6) into (2)
k+16 = 25
Therefore k=9
2007-01-09 8:55 am
X^2 + 16 = 5X + Y .....(1)
k + Y = 5X .......(2)
From (2), we have Y = 5X - k.
Putting it into (1):
X^2 + 16 = 5X + (5X - k)
X^2 - 10X + (16 + k) = 0
As the set of simutaneous equations have one unique solution,
distriminant = 0
(-10)^2 - 4 (1) (16 + k) = 0
k = 100/4 - 16
k = 9.
k + Y = 5X .......(2)
From (2), we have Y = 5X - k.
Putting it into (1):
X^2 + 16 = 5X + (5X - k)
X^2 - 10X + (16 + k) = 0
As the set of simutaneous equations have one unique solution,
distriminant = 0
(-10)^2 - 4 (1) (16 + k) = 0
k = 100/4 - 16
k = 9.
參考: 出處: myself
2007-01-09 1:47 am
put the 2nd equation into the 1st one
X^2 + 16 = (K + Y) + Y
X^2 + 16 = K + 2Y
K = X^2 - 2Y +16
X^2 + 16 = (K + Y) + Y
X^2 + 16 = K + 2Y
K = X^2 - 2Y +16
收錄日期: 2021-05-02 14:11:38
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