Question

helpppp guyssss helppppppppppp​

Answer 1

Step-by-step explanation:

$\frac{1}{(x - 1)(x - 2)} + \frac{1}{(x - 2)(x - 3)} = \frac{2}{3} \\ \\ take \: LCM \: in \: LHS \\ \\ \frac{x - 3 + x - 1}{(x - 1)(x - 2)(x - 3)} = \frac{2}{3} \\ \\ \frac{2x - 4}{( {x}^{2} - 3x + 2)(x - 3) } = \frac{2}{3} \\ \\ \frac{2(x - 2)}{ {x}^{3} - 3 {x}^{2} - 3 {x}^{2} + 9x + 2x - 6 } = \frac{2}{3} \\ \\ cancel \: 2 \: from \: both \: sides \\ \\ \frac{x - 2}{ {x}^{3} - 6 {x}^{2} + 11x - 6 } = \frac{1}{3} \\ \\ cross \: multiply \\ \\ 3(x - 2) = {x}^{3} - 6 {x}^{2} + 11x - 6 \\ \\ {x}^{3} - 6 {x}^{2} + 11x - 6 - 3x + 6 = 0 \\ \\ {x}^{3} - 6 {x}^{2} + 8x = 0 \\ \\ x( {x}^{2} - 6x + 8) = 0 \\ \\ x = 0 \\ \\ or \\ \\ {x}^{2} - 6x + 8 = 0 \\ \\ {x}^{2} - 4x - 2x + 8 = 0 \\ \\ x(x - 4) - 2(x - 4) = 0 \\ \\ (x - 4)(x - 2) = 0 \\ \\ x - 4 = 0 \\ \\ x = 4 \\ \\ or \\ \\ x - 2 = 0 \\ \\ x = 2$

Values of x can be 0,2 and 4,but in the question it is given that x≠1,2,3

So,discard x=2

Value of x: 0 and 4

Hope you understand

Answer 2

Given :   1/(x - 1)(x - 2)   +  1/(x - 2)(x - 3)   = 2/3  , x  ≠ 1 , 2 , 3

To find : Value of x  

Solution:

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

multiplying both sides by  (x - 1)(x - 3)

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

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

=> (2x - 4)/(x - 2) = 2 (x - 1)(x - 3)/3

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

=>  2 = 2 (x - 1)(x - 3)/3

=> 1  = (x - 1)(x - 3)/3

=>  (x - 1)(x - 3) = 3

=> x² -  4x  + 3 = 3

=> x² - 4x = 0

=> x(x - 4) = 0

=> x = 0  , 4

x  =  0   &  4

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