maths(解方程 二 : 代數方法)

2. 可以,但留意是
一. 開方內必要是正,本題即x-1 > 0或x > 1
二. 有時因作平方會加了一個解,如第四題,所以解後要覆實答案.
3. √(x - 1) - 3 = 0
√(x - 1) = 3
x - 1 = 9
x = 10
4. √(x + 3) - x = 1
√(x + 3) = 1 + x
x + 3 = 1 + 2x + x2
x2 + x - 2 = 0
(x - 1)(x + 2) = 0
x = 1或 x= -2
但√(-2 + 3) - (-2) = 3
所以x = 1
1. x - (3 / x + 1) = 1
如果是 x - (3/x + 1) = 1
則 x2 - 3 - x = x
x2 - 2x - 3 = 0
(x - 3)(x + 1) = 0
x = 3 或x = -1
如果是 x - 3/(x + 1) = 1
x(x + 1) - 3 = x + 1
x2 + x - 3 = x + 1
x2 - 4 = 0
(x - 2)(x + 2) = 0
x = 2或x = -2
2. 9x4 - 37x2 + 4 = 0
=> 9x4 - 12x2 + 4 - 25x2 = 0
=> (3x2 - 2)2 - (5x)2 = 0
=> (3x2 + 5x - 2)(3x2 - 5x - 2) = 0
=> (3x - 1)(x + 2)(3x + 1)(x - 2) = 0
=> x = -2, -1/3, 1/3 或 2
3. 2x - √(2x) = 0
=> √(2x)[√(2x) - 1] = 0
=> √(2x) = 0 或 √(2x) = 1
=> x = 0 或 x = 0.5
4. 22x - 4(2x) + 3 = 0
Let u = 2x
u2 - 4u + 3 = 0
=> (u - 3)(u - 1) = 0
=> u = 3 或 u = 1
=> 2x = 3 或 2x = 1
=> x = log3/log2 或 x = 0