Question

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

Answer 1

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

    1    [     1     +     1     ]  =  2 -------------[Taken 1 / (x - 2) Common ]
-------   ---------    ---------      ----
(x - 2)   (x - 1)    (x  -  3)       3
   
   1     [x - 3 + x - 1]  = 2 
-------  ------------------  ----
(x - 2)  (x - 1)(x  -  3)   3
   
   1     [   2x - 4   ]   = 2 
-------   -------------     ----
(x - 2)  x² - 4x + 3      3
 
        2x - 4                   =  2
----------------------------     -----
x³ - 4x² - 2x²  + 3x - 6       3

        2x - 4             =  2
----------------------      -----
x³ - 6x² + 3x - 6         3

(2x - 4)3 = 2(x³ - 6x² + 3x - 6)
6x -12 = 2x³ -12x² +6x -12--[6x - 12 gets cancelled on both the sides]
2x³ - 12x² = 0
2x²( x - 6 ) = 0

2x² = 0  OR x - 6 = 0
x = 0     OR  x = 6

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Answer 2

$\huge{\mathcal{Hi\: there!}}$

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$\huge{\bold{Solution :}}$

$\mathsf{ \large{ \frac{x - 3 + x - 1}{(x - 1)(x - 2)(x - 3)} = \frac{2}{3}}} \\ \\ \mathsf{ \frac{2x - 4}{x {}^{3} -6x {}^{2} + 11x - 6} = \frac{2}{3} } \\ \\ \mathsf{\frac{2(x - 2)}{x{}^{3} -6x {}^{2} + 11x - 6} = \frac{2}{3} } \\ \\ \mathsf{ \frac{x - 2}{x{}^{3} -6x {}^{2} + 11x - 6} = \frac{1}{3} } \\ \\ \mathsf{ 3x - 6 = x{}^{3} -6x {}^{2} + 11x - 6} \\ \\ \mathsf{{}x^{3} -6x {}^{2} + 8x = 0} \\ \\ \mathsf{x(x {}^{2} - 6x + 8) = 0 } \\ \\ \mathsf{x {}^{2} - 6x + 8 = 0}$

Now, solve this equation, to get the value of x ;

$\mathsf{x {}^{2} - 6x + 8 = 0} \\ \\ \mathsf{x {}^{2} - 4x - 2x + 8 = 0 } \\ \\ \mathsf{x(x - 4) - 2(x - 4) = 0} \\ \\ \mathsf{(x - 2)(x - 4) = 0}$

Hence,

$\mathsf{(x - 2) = 0} \\ \\ \therefore\mathsf{x = 2}$

And,

$\mathsf{(x - 4) = 1} \\ \\ \therefore \mathsf{x = 4}$

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