Question

1) 1/(x - 1)(x - 2) + 1/(x - 2)(x - 4) + 1/(x - 3)(x - 4) = 1/6 where, x not equal to 1,2,3,4.
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Answer 1

EXPLANATION.

⇒ 1/(x - 1)(x - 2) + 1/(x - 2)(x - 4) + 1/(x - 3)(x - 4) = 1/6.

Taking L.C.M in this equation, we get.

⇒ (x - 3)(x - 4) + (x - 1)(x - 4) + (x - 1)(x - 2)/(x - 1)(x - 2)(x - 3)(x - 4) = 1/6.

⇒ (x - 3)(x - 4) + (x - 1)[(x - 4) + (x - 2)]/(x - 1)(x - 2)(x - 3)(x - 4) = 1/6.

⇒ (x - 3)(x - 4) + (x - 1)[x - 4 + x - 2]/(x - 1)(x - 2)(x - 3)(x - 4) = 1/6.

⇒ (x - 3)(x - 4) + (x - 1)(2x - 6)/(x - 1)(x - 2)(x - 3)(x - 4) = 1/6.

⇒ (x - 3)(x - 4) + (x - 1)2(x - 3)/(x - 1)(x - 2)(x - 3)(x - 4) = 1/6.

⇒ (x - 3)[(x - 4) + 2(x - 1)]/(x - 1)(x - 2)(x - 3)(x - 4) = 1/6.

⇒ (x - 3)[x - 4 + 2x - 2]/(x - 1)(x - 2)(x - 3)(x - 4) = 1/6.

⇒ (x - 3)(3x - 6)/(x - 1)(x - 2)(x - 3)(x - 4) = 1/6.

⇒ (x - 3)3(x - 2)/(x - 1)(x - 2)(x - 3)(x - 4) = 1/6.

⇒ 3/(x - 1)(x - 4) = 1/6.

⇒ (x - 1)(x - 4) = 18.

⇒ x² - 4x - x + 4 = 18.

⇒ x² - 5x + 4 = 18.

⇒ x² - 5x - 14 = 0.

Factorizes the equation into middle term split, we get.

⇒ x² - 7x + 2x - 14 = 0.

⇒ x(x - 7) + 2( x - 7) = 0.

⇒ (x + 2)(x - 7) = 0.

⇒ x = -2  & x = 7.

Answer 2

Answer:

x = -2 and 7

Step-by-step explanation:

Correct Question:

$\bf\dfrac{1}{(x - 1)(x - 2)}$ + $\bf\dfrac{1}{(x - 2)(x - 3)}$ + $\bf\dfrac{1}{(x - 3)(x - 4)}$ = $\bf\dfrac{1}{6}$ where, x not equal to 1,2,3,4

To find:

The value of x

Solution:

Now, let us take LCM:

$\bf\dfrac{(x-3)(x-4)+(x-1)(x-4)+(x-1)(x-2)}{(x-1)(x-4)(x-2)(x-3)}$

$\bf\dfrac{(x^2-4x-3x+12)+(x^2-4x-x+4)+(x^2-2x-x+2)}{(x^2-5x+4)(x^2-5x+6)}$

$\bf\dfrac{(x^2-7x+12)+(x^2-5x+4)+(x^2-3x+2)}{(x^2-5x+4)(x^2-5x+6)}$

$\bf\dfrac{3x^2-15x+18}{(x^2-5x+4)(x^2-5x+6)}$

Take 3 as common here:

$\bf\dfrac{3(x^2-5x+6)}{(x^2-5x+4)(x^2-5x+6)}$

$\bf{x^2-5x+6}$ gets cancelled on dividing:

$\bf\dfrac{3}{x^2-5x+4}$ = $\bf\dfrac{1}{6}$

$\bf{x^2-5x+4 = 18}$

$\bf{x^2-5x-14 = 0}$

$\bf{x^2-7x+2x-14 = 0}$

Take x and 2 as common respectively:

$\bf{x(x-7)+2(x-7) = 0}$

$\bf{(x+2)(x-7)}$

x+2 = 0 and x-7 = 0

x = -2 and 7