當y=-1時
y=(7/3)x+(5/3)/x
數學問題呀~
2006-12-23 6:45 am
回答 (4)
2006-12-23 6:48 am
✔ 最佳答案
y=(7/3)x+(5/3)/x when y=-1
-1= (7/3)x+(5/3)x
-1= (12/3)x
-1=4x
x=-1/4
要通分母ａｒ
參考: me
2006-12-23 7:42 am
no real roots
2006-12-23 7:31 am
y = (7/3) x + (5/3) / x
當 y = -1,
-1 = (7/3) x + (5/3) / x
-3x = 7x^2 + 5 [左右乘3x]
7x^2 + 3x + 5 = 0
Δ = b^2 - 4ac
= 3^2 - 4(7)(5)
= -131
< 0
所以沒有實根
當 y = -1,
-1 = (7/3) x + (5/3) / x
-3x = 7x^2 + 5 [左右乘3x]
7x^2 + 3x + 5 = 0
Δ = b^2 - 4ac
= 3^2 - 4(7)(5)
= -131
< 0
所以沒有實根
2006-12-23 6:57 am
當y=-1的時候
-1 = (7/3)X + (5/3)X
-1 = (7X/3) + (5X/3)
-1 = (7X + 5X)/3
-3 = 12X
(-3/12) = X
X = -1/4
-1 = (7/3)X + (5/3)X
-1 = (7X/3) + (5X/3)
-1 = (7X + 5X)/3
-3 = 12X
(-3/12) = X
X = -1/4
收錄日期: 2021-04-12 22:46:49
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