f.4數學題(二次方程)

第一題...

3x^4-8x^2-3=0
(3x^2 + 1)(x^2 - 3)=0
x^2 = - 1/3 (rejected) or x^2 = 3

so x = sq. root of 3 (即係開方3)

第二題...如果我冇估錯...
應該係...
1/(x+2)+3/[2(x+2)]=(2x+1)/8

咁...
1/(x+2)+3/[2(x+2)]=(2x+1)/8
{1/(x+2)+3/[2(x+2)]}(x+2)=[(2x+1)/8](x+2)
1 + 3/2 = (2x + 1)(x+2)/8
1 + 3/2 = (2x^2 + 5x + 2)/8
8 + 12 = 2x^2 + 5x + 2
2x^2 + 5x - 18 = 0
(2x + 9)(x - 2) = 0
x = -4.5 or x = 2

第三條...

x/5(x-3)-3/2(x-3)=x-1/2
x/[5(x-3)] - 3/[2(x-3)] = x - 1/2
{x/[5(x-3)] - 3/[2(x-3)]}[10(x-3)] = (x - 1/2)[10(x-3)]
2x - 15 = 10x^2 - 35x + 15
10x^2 - 37x + 30 = 0
(2x - 5)(5x - 6) = 0
x = 2.5 or x = 1.2

第四條...

(x+10)(3-4/x)=12
(x+10)[(3x/x)-(4/x)]=12
(x+10)[(3x-4)/x]=12
{(x+10)[(3x-4)/x]}(x)=12(x)
(x+10)(3x-4) = 12x
3x^2 +26x -40 = 12x
3x^2 + 14x - 40 = 0
(3x + 20)(x - 2) = 0
x = - 20/3 or x = 2

^^

上面果位人兄第一題好似計錯左

1. 3x^4 - 8x² - 3 = 0

Let y = x²
3y² - 8y - 3 = 0
(3y + 1 )( y - 3 ) = 0
y = -1/3 or 3
y = x² = -1/3(rejected) y = x² = 3
x = +√3 or -√3

2006-10-15 17:42:25 補充：
小學都有教啦 x² = y, x= 正or負的 √y咁多人上黎亂答 証明係度問問題係好唔可靠

1.
3x^4-8x²-3=0
(x²-3)(3x²+1)=0
x=√3 或 ±√-1（捨去）

2.
1/x+2+3/2(x+2)=2x+1/8
8x(x+2)[1/x+2+3/2(x+2)]=8x(x+2)(2x+1/8)
8(x+2)+16x(x+2)+12x=16(x+2)+x(x+2)
8x+16+16x²+32x+12x=16x+32+x²+2x
15x²+34x-16=0
(5x-2)(3x+8)=0
x=2/5 或 -8/3

3.
x/5(x-3)-3/2(x-3)=x-1/2
10(x-3)[x/5(x-3)-3/2(x-3)]=10(x-3)(x-1/2)
2x-15=5(x-3)(2x-1)
2x-15=5(2x²-7x+3)
2x-15=10x²-35x+15
10x²-37x+30=0
(2x-5)(5x-6)=0
x=5/2 或 6/5

4.
(x+10)(3-4/x)=12
x(x+10)(3-4/x)=12x
(x+10)(3x-4)=12x
3x²+26x-40=12x
3x²+14x-40=0
(x-2)(3x+20)=0
x=2 或 -20/3