Solve for 1/(x-1)(x-2) + 1/(x-3)(x-2) = 2/3, x ≠ 1, 2, 3.

Solution:

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

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

3[(x-3) + (x-1)] = 2(x-1)(x-2)(x-3) 

3(2x – 4) = 2(x-1)(x-2)(x-3) 

3×2(x-2) = 2(x-1)(x-2)(x-3) 

3 = (x-1)(x-3) 

(x2– 4x + 3) – 3 = 0

=> x2 – 4x = 0

=>x(x-4) = 0

=> x = 0 or x = 4

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