Question

plz find the value of x....

Answer 1

to solve for x.
We know,  x ≠1 , x≠2, x≠3, x≠4.

$\frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}=\frac{1}{6}-\frac{1}{(x-3)(x-4)}\\\\\frac{x-3+x-1}{(x-1)(x-2)(x-3)}=\frac{x^2-7x+12-6}{6(x-3)(x-4)}\\\\\frac{2(x-2)}{(x-1)(x-2)(x-3)}=\frac{x^2-7x+6}{6(x-3)(x-4)}\\\\12(x-4)=(x-1)(x^2-7x+6)\\\\x^3-8x^2+x+42=0$

x³-8x²+x+42=0
     Checking for x=3,-3, -4, 4, -1, we get x =3 works out.

So (x-3) (x² - 5x -14 ) = 0
      as  x ≠3,     (x-7) (x+2) = 0

Hence,   x = 7  or   -2...   
Check the solution. That is correct.