標題:

Quadratic Equations, 高手請進

發問:

How to solve: 1. [4/(x+2)]-4=3/(x+3) 2. 1/(x+1)=(x-1)/(x+5)

最佳解答:

Solution 1.[4/(x+2)]-4=3/(x+3)(x≠-2, x≠-3) [4/(x+2)]-4(x+2)/(x+2)=3/(x+3) [4 -4(x+2)]/(x+2) =3/(x+3) [4 -4x - 8]/(x+2) =3/(x+3) (-4x – 4)/(x+2) =3/(x+3) (-4x – 4) (x+3) =3(x+2) -4x^2 – 4x – 12x -12 = 3x + 6 0 = 3x + 6 +4x^2 + 4x + 12x +12 0 = 4x^2 + 19x +18 4x^2 + 19x +18 = 0 x = {(-19) ±√ [19^2 – 4 (4)(18)]}/[(2)(4)] x = [-19 ±√ (361 – 288)]/8 x = [-19 ±√ (361 – 288)]/8 x = (-19 ±√73)/8 x = (-19 +√73)/8 or x = (-19 -√73)/8 x = -1.307 or x = -3.443 2. 1/(x+1)=(x-1)/(x+5) (x≠-1, x≠-5) 1(x+5) =(x-1) (x+1) x + 5 = x^2 – 1 x^2 – x – 1 – 5 = 0 x^2 – x – 6 = 0 (x + 2) (x – 3) = 0 x = -2 or x = 3

其他解答:

此文章來自奇摩知識+如有不便請留言告知

1. [4/(x+2)]-4=3/(x+3) 4(x+3) - 4(x+3)(x+2) = 3(x+2) . [multiply both side by (x+3)(x+2)] 4x+12 - 4(x^2+5x+6) = 3x+6 4x^2+19x-18=0 x=-1.307 or -3.443 2. 1/(x+1)=(x-1)/(x+5) x+5 = (x-1)(x+1) [multiply both side by (x+1)(x+5)] x+5 = x^2 - 1 x^2 - x -6 = 0 (x-3)(x+2) = 0 x=3 or -2 2006-10-17 09:52:05 補充： typing error, the fourth line should be : 4x^2 19x 18=0 2006-10-17 09:53:09 補充： typing error again, the fourth line should be : 4x^2 ┼ 19x ┼ 18=0|||||1) 4/(x+2) - 3/(x+3) =4 4(x+3)-3(x+2)=4(x+2)(x+3) x+6=4(x^2+19x+6) 4X^2+19x+18=0 x=-1.307 or x=-3.443 2) 1/(x+1)= (x-1)/(x+5) x+5 = (x+1)(x-1) x+5=x^2-1 x^2 -x -6 =0 x=-2 or x=3FAD2A23AB937987B

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