Question

find x
$\frac{1}{(x - 1)(x - 2)} + \frac{1}{(x - 2)(x - 3)} = \frac{2}{3}$
x ≠ 1 , 2 , 3 ​

Answer 1

ANSWER:

Given:

• 1/(x-1)(x-2) + 1/(x-2)(x-3) = 2/3

To Find:

• Value of x

Solution:

$\text{We are given that,}\\\\:\longrightarrow\dfrac{1}{(x-1)(x-2)}+\dfrac{1}{(x-2)(x-3)}=\dfrac{2}{3}\\\\\text{Taking LCM,}\\\\:\implies\dfrac{(x-3)+(x-1)}{(x-1)(x-2)(x-3)}=\dfrac{2}{3}\\\\:\implies\dfrac{x-3+x-1}{(x-1)(x-2)(x-3)}=\dfrac{2}{3}\\\\:\implies\dfrac{2x-4}{(x-1)(x-2)(x-3)}=\dfrac{2}{3}\\\\:\implies\dfrac{2(x-2)}{(x-1)(x-2)(x-3)}=\dfrac{2}{3}\\\\\text{(x-2) gets cancelled,}\\\\:\implies\dfrac{2\!\!\!/}{(x-1)(x-3)}=\dfrac{2\!\!\!/}{3}\\\\:\implies\dfrac{1}{(x-1)(x-3)}=\dfrac{1}{3}\\\\\text{On cross-multiplying,}\\\\:\implies3=(x-1)(x-3)\\\\:\implies3\!\!\!/\:=x^2-4x+3\!\!\!/\\\\:\implies x^2-4x=0\\\\:\implies x(x-4)=0\\\\:\implies x-4=0\:or\:x=0\\\\\bf{:\implies x=4\:and\:0}$

Verification:

$:\longrightarrow\dfrac{1}{(x-1)(x-2)}+\dfrac{1}{(x-2)(x-3)}=\dfrac{2}{3}\\\\\text{Solving LHS, at x=4,}\\\\:\implies\dfrac{1}{(4-1)(4-2)}+\dfrac{1}{(4-2)(4-3)}\\\\:\implies\dfrac{1}{3\times2}+\dfrac{1}{2\times1}\\\\:\implies\dfrac{1}{6}+\dfrac{1}{2}\\\\\text{Taking LCM,}\\\\:\implies\dfrac{1+3}{6}\\\\:\implies\dfrac{4\!\!\1/^{\:2}}{6\!\!\!/_{\:3}}\\\\\bf{:\implies\dfrac{2}{3} = RHS}- - - -(1)$

$\text{Solving LHS, at x=0,}\\\\:\implies\dfrac{1}{(0-1)(0-2)}+\dfrac{1}{(0-2)(0-3)}\\\\:\implies\dfrac{1}{-1\times-2}+\dfrac{1}{-2\times-3}\\\\:\implies\dfrac{1}{2}+\dfrac{1}{6}\\\\\text{Taking LCM,}\\\\:\implies\dfrac{3+1}{6}\\\\:\implies\dfrac{4\!\!\1/^{\:2}}{6\!\!\!/_{\:3}}\\\\\bf{:\implies\dfrac{2}{3} = RHS}- - - -(2)\\\\\text{From (1) & (2),}\\\\\text{\bf{HENCE VERIFIED!!!}}$

Answer 2

best answer ke liye refer the attachment.

Agar achha lage to like zarur krna.