x^+7/2x-2=0的解

您好，我是 lop，高興能解答您的問題。

你寫得很不清楚，我分開了幾個 cases：

Case 1 :

x^2 + (7/2)x - 2= 0
2[x^2-(7/2)x-2] = 0
2x^2 - 7x - 4 = 0
( 2x^2 - 8x ) + ( x - 4 ) = 0
2x(x-4) + 1(x-4) = 0
(2x+1)(x-4) = 0
2x+1=0 或 x-4=0
x = -0.5 或 4

Case 2 :

x^2 + 7/(2x) - 2 = 0
2x[x^2+7/(2x)-2] = 0
2x^3 - 4x + 7 = 0
( http://www.wolframalpha.com/input/?i=2x%5E3+-+4x+%2B+7+%3D+0 )

Case 3 :

x + (7/2)x - 2 = 0
2[x+(7/2)x-2] = 0
2x + 7x - 2 = 0
9x - 2 = 0
9x - 2 + 2 = 0 + 2
9x = 2
9x/9 = 2/9
x = 2/9

Case 4 :

x + 7/(2x) - 2 = 0
2x[x+7/(2x)-2] = 0
2x^2 - 4x + 7 = 0
( http://www.wolframalpha.com/input/?i=2x%5E2+-+4x+%2B+7+%3D+0 )

Case 5 :

x + 7/(2x-2) = 0
(2x-2)[x+7/(2x-2)] = 0
x(2x-2) + 7 = 0
2x^2 - 2x + 7 = 0
( http://www.wolframalpha.com/input/?i=2x%5E2+-+2x+%2B+7+%3D+0 )

Case 6 :

x^2 + 7/(2x-2) = 0
(2x-2)[x^2+7/(2x-2)] = 0
(2x-2)x^2 + 7 = 0
2x^3 - 2x^2 + 7 = 0
( http://www.wolframalpha.com/input/?i=2x%5E3+-+2x%5E2+%2B+7+%3D+0 )

Case 7 :

(x^7)/(2x) - 2 = 0
[x(x^6)]/[x(2)] - 2 = 0
(x^6)/2 - 2 = 0
2[(x^6)/2-2] = 0
x^6 - 4 = 0
(x^3)^2 - 2^2 = 0
(x^3+2)(x^3-2) = 0
( http://www.wolframalpha.com/input/?i=%28x%5E3%2B2%29%28x%5E3-2%29+%3D+0 )

Case 8 :

(x^7)/(2x-2) = 0
(2x-2)[(x^7)/(2x-2)] = 0
x^7 = 0
x = 0

P.S. 但我想你都係指 Case 1。

2011-10-02 17:46:06 補充：
corr.

Case 1 :

x^2 + (7/2)x - 2= 0
2[x^2+(7/2)x-2] = 0
2x^2 + 7x - 4 = 0
( 2x^2 + 8x ) - ( x + 4 ) = 0
2x(x+4) - 1(x+4) = 0
(2x-1)(x+4) = 0
2x-1=0 或 x+4=0
x = -4 或 0.5

2011-10-05 11:43:12 補充：
x^2 + (7/2)x - 2 = 0
兩根和 = -b = -7/2
兩根積 = c = -2

x^2+7/2x-2=0
2(x^2+7/2x-2)=(2)0
2x^2+7x-4=0
(2x-1)(x+4)=0
2x-1=0 or x+4=0
2x=1 or x=-4
x=1/2 or x=-4

一般式是ax^2+bx+c=0 (a不=0)

x^+7/2x-2=0???
我估你係想講x^2+7/2x-2=0
x^2+7/2x-2=0
2x^2+7x-4=0x2
2x^2+7x-4=(2x-1)(x+4)=0
x=0.5,-4

x^2+7/2x-2=0
2x^2+7x-4=0
(2x-1)(x+4)=0
x=1/2 or x=-4