Question

pls say this answer​

Answer 1

Answer:

dontknowfollowmeifyourfromclass8

Answer 2

Answer:

Though we can proceed by taking LCM of denominators and trying to solve the fourth degree equation we will get. That will be troublesome for us. Instead we try breaking every term as sum/difference of two terms as given below:

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

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

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

adding above we get

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

hence our problem becomes

1/(x-4) - 1)(x-1) = 1/6

(x - 1 - x + 4)/(x-1)(x-4) = 1/6

(x-1)(x-4) = 18

x^2 - 5x + 4 - 18 = 0

x^2 - 5x - 14 = 0

x^2 - 7x + 2x - 14 = 0

x(x-7) + 2(x-7) = 0

(x-7) (x+2) = 0

x = 7, x = -2

in this regard, it is worthwhile to mention that originally the equation would be a fourth degree equation so ideally we should have four solutions. however to avoid complexity, we proceeded to find two solutions only