`急......................F.4Math Quadratic Equations

1)
[1/(x+1)] + [1/(x-1)] = 3/4
Multiply all terms on the both sides by 4(x+1)(x-1):
4(x+1)(x-1)[1/(x+1)] + 4(x+1)(x-1)[1/(x-1)] = 4(x+1)(x-1)(3/4)
4(x-1)+4(x+1) = 3(x+1)(x-1)
4x- 4+4x+4 = 3x2-3
3x2-8x-3 = 0
(3x+1)(x-3) = 0
x = -1/3 or x = 3

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2)
[(12/x)+1][2-(3/x)] = 5
Multiply all terms on the both sides by x2.
[(12/x)+1]xŸ[2-(3/x)]x = 5x2
(12+x)(2x-3) = 5x2
24x-36+2x2-3x = 5x2
3x2-21x+36 = 0
x2-7x+12=0
(x-3)(x=4) = 0
x = 3 or x = 4

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3)(a)
x = 3+2√(2x-1) ...... (1)

Let u = √(2x - 1) ...... (3)
u2 = [√(2x-1)]2
u2 = 2x-1
2x = u2+1
x = (u2+1)/2 ...... (4)

Put (3) and (4) into (1)
(u2+1)/2 = 3+2u
u2+1 = 2(3+2u)
u2+1 = 6+4u
u2-4u-5 = 0 ...... (2)

3)(b)
(2):
u2-4u-5 = 0
(u-5)(u+1) = 0
u = 5 or u = -1

(3):
u = √(2x - 1)

When u = -1
√(2x - 1) = -1 (rejected)

When u = 5
√(2x - 1) = 5
2x - 1 = 52
2x -1 = 25
2x = 26
x = 13
=

1) (1/(x+1))+(1/(x-1))=3/4
4[(x-1)+(x+1)] = 3(x-1)(x+1)
8x = 3x2-3
3x2-8x-3 = 0
(3x+1)(x-3) = 0
x = -1/3 or x = 3

2) ((12/x)+1)(2-(3/x))=5
[(12+x)/x][(2x-3)/x] = 5
(12+x)(2x-3) = 5x2
2x2+21x-36 = 5x2
3x2-21x+36 = 0
x2-7x+12 = 0
(x-3)(x-4) = 0
x = 3 or x = 4

3) u2-4u-5
= (√2x-1)2-4(√2x-1)-5
= 2x2-2√2x+1-4√2x+4-5
= 2x2-6√2x
= 2x(x-3√2)