Solve the equation...?

Assuming your question is:

2(x - 2)/(x^2 - 4) + 3/(2x - 1) = 1
2(x - 2)/[(x + 2)(x - 2)] + 3/(2x - 1) = 1
2/(x + 2) + 3/(2x - 1) = 1

Common factor is (x + 2)(2x -1)

2(2x - 1) + 3(x + 2) = (x + 2)(2x - 1)
4x - 2 + 3x + 6 = 2x^2 - x + 4x - 2
7x + 4 = 2x^2 + 3x -2
2x^2 - 4x - 6 = 0
(2x + 2)(x - 3) = 0

Therefore,
2x + 2 = 0 which gives x = -1
and x - 3 = 0 which gives x = 3.

(2[x - 2]/[x² - 4]) + (3/[2x - 1]) = 1
(2[x - 2]/[{x - 2}{x + 2}]) + (3/[2x - 1]) = 1
(2/[x + 2]) + (3/[2x - 1]) = 1
(4x - 2 + 3x + 6)/([x + 2][2x - 1]) = 1
(7x + 4)/(2x² - x + 4x - 2) = 1
7x + 4 = 2x² + 3x - 2
2x² - 4x = 6
x² - 2x = 3
x² - x = 3 + 1²
x² - x = 3 + 1
(x - 1)² = 4
x - 1 = 2

x = 2 + 1, x = 3
x = - 2 + 1, x = - 1

Answer: x = 3, - 1; proofs are given below.

(2[3 - 2]/[3² - 4]) + (3/[2{3} - 1]) = 1
(2[1]/[9 - 4]) + (3/[6 - 1]) = 1
2/5 + 3/5 = 1
5/5 = 1
1 = 1

(2[- 1 - 2]/[- 1² - 4]) + (3/[2{- 1} - 1]) = 1
(2[- 3]/[1 - 4]) + (3/[- 2 - 1]) = 1
(- 6/- 3) + (3/- 3) = 1
2 + (- 1) = 1
2 - 1 = 1
1 = 1

2(x - 2)/(x^2 - 4) + 3/(2x - 1) = 1
2(x - 2)/(x^2 - 2^2) + 3/(2x - 1) = 1
2(x - 2)/(x + 2)(x - 2) + 3/(2x - 1) = 1
2/(x + 2) + 3/(2x - 1) = 1
[x + 2][2x - 1][2/(x + 2) + 3/(2x - 1)] = [x + 2][2x - 1]
2(2x - 1) + 3(x + 2) = (x)(2x) + (2)(2x) - (x)(1) + (2)(-1)
4x - 2 + 3x + 6 = 2x^2 + 4x - x + (-2)
2x^2 + 3x - 4x - 3x - 2 + 2 - 6 = 0
2x^2 - 4x - 6 = 0
(2x^2 - 4x - 6)/2 = 0/2
x^2 - 2x - 3 = 0
x^2 + x - 3x - 3 = 0
(x^2 + x) - (3x + 3) = 0
x(x + 1) - 3(x + 1) = 0
(x + 1)(x - 3) = 0

x + 1 = 0
x = -1

x - 3 = 0
x = 3

∴ x = -1 , 3

2 (x - 2) / [ (x - 2)(x + 2) ] + 3 / (2x - 1) = 1

2 / [ (x + 2) ] + 3 / (2x - 1) = 1

2 (2x - 1) + 3 (x + 2) = (x + 2)(2x - 1)

4x - 2 + 3x + 6 = 2x² + 3x - 2

0 = 2x² - 4x - 6 = 0

x² - 2x - 3 = 0

(x - 3)(x + 1) = 0

x = 3 , x = - 1