二次方程..

1)1÷x^2-2÷x-7 = 0
1 - 2x -7x^2 = 0
7x^2 +2x -1 = 0
x = (-2正負開方32)/2
= -1 正負開方8

2007-09-19 20:12:08 補充：
1) 1÷x^2-2÷x-7 = 0兩邊乘 x ^ 21 - 2x^2/x - 7x^2 = 07x^2 2x -1 =0a =7,b = 2 c = -1x = [-2正負開方(2^2 28)]/2x = -1 正負開方82)x 1÷x = 29÷10兩邊乘 xx^2 - 29x/10 1 = 010x^2 - 29x 10 =0(2x 5)(5x 2) =0x = - 2/5 or x = - 5/2

1)
(1÷x^2-2÷x-7)[x^2(x-7)] = 0[x^2(x-7)]
1(x-7)-2(x^2) = 0
-2x^2+x-7 = 0
Δ=(1)^2-4(-2)(-7)=-55 < 0
∴此題無解。

2)
x+1÷x = 29÷10-------(1)
x(x+1÷x) = x(29÷10)
x^2+1 = 2.9x
x^2-2.9x+1 = 0

利用2次公式:
x={-(-2.9)±√[(-2.9)^2 - 4(1)(1)]} / [2(1)] = 0.4 或 2.5
(√即開方)

代入x=0.4入(1):
左式=2.9
右式=2.9
∴x=0.4正確

代入x=2.5入(1):
左式=2.9
右式=2.9
∴x=2.5正確

∴x = 0.4 或 2.5

1) 1÷x^2-2÷x-7 = 0
這題目有問題，没有一個數被一除會等如零
但如果題目是：
1/[ (x^2-2)/(x-7)] =0
因為1/[ (x^2-2)/(x-7)] = (x-7)/(x^2-2)
所以如果1/[ (x^2-2)/(x-7)] =0
(x-7)/(x^2-2)=0
x-7=0
x=7
2) x+1÷x = 29÷10
(x+1)/x = 29/10
交叉相乘
29x= 10(x+1)
29x= 10x + 10
19x= 10
x= 10/19