Question

Attention!!!⚠️
Answer this clearly
??​

Answer 1

Question:

Solve for x :-

$\frac{1}{(x - 1)(x - 2)} + \frac{1}{(x - 2)(x - 3)} = \frac{2}{3}$

Answer:

x = 0 , 4

Solution:

We have;

$= > \frac{1}{(x - 1)(x - 2)} + \frac{1}{(x - 2)(x - 3)} = \frac{2}{3}$

$= > \frac{1}{(x - 1)(x - 2)} + \frac{1}{(x - 2)(x - 3)} - \frac{2}{3} = 0$

$= > \frac{3(x - 3) + 3(x - 1) - 2(x - 1)(x - 2)(x - 3)}{(x - 1)(x - 2)(x - 3)} = 0$

$= > 3(x - 3) + 3(x - 1) - 2(x - 1)(x - 2)(x - 3) = 0 \times (x - 1)(x - 2)(x - 3)$

$= > 3(x - 3) + 3(x - 1) - 2(x - 1)(x - 2)(x - 3) = 0$

$= > 3(x - 3 + x - 1) - 2(x - 1)(x - 2)(x - 3) = 0$

$= > 3(2x - 4) - 2(x - 1)(x - 2)(x - 3) = 0$

$= > 3 \times 2(x - 2) - 2(x - 1)(x - 2)(x - 3) = 0$

$= > 2 \times ( \: 3(x - 2) - (x - 1)(x - 2)(x - 3) \: ) = 0$

$= > 3(x - 2) - (x - 1)(x - 2)(x - 3) = 0$

$= >(x - 2)( \:3 - (x - 1)(x - 3) \: ) = 0$

$= >(x - 2)( \:3 - ( {x}^{2} - 4x + 3) \: ) = 0$

$= >(x - 2)( \:3 - {x}^{2} + 4x - 3 \: ) = 0$

$= >(x - 2)( - {x}^{2} + 4x) = 0$

$= > (x - 2)( - x)(x - 4) = 0$

$= > x(x - 2)(x - 4) = 0$

$= > \: \: x = 0 \: \: or \: \: x - 2 = 0 \: \: or \: \: x - 4 = 0$

$= > x = 0 \: \: or \: \: x = 2 \: \: or \: \: x = 4$

But ,

It is given that ;

x ≠ 2

Hence,

The appropriate values of x are : 0 , 4 .

Answer 2

Refer the attachment.

Answer = x = 0, 4