We have to simplify 2x/(x^2 -6x+9) - 1/(x+1) - 8/(x^2 -2x-3)

2x/(x^2 -6x+9)

=> 2x/(x^2 -3x - 3x+9)

=> 2x/ [x( x - 3) - 3(x- 3)]

=> 2x/ (x-3)^2

8/ ( x^2 - 2x - 3)

=> 8/( x^2 - 3x + x - 3)

=> 8/[ x(x - 3) + 1( x - 3)

=> 8/ ( x+1)(x-3)

So 2x/(x^2 -6x+9) - 1/(x+1) - 8/(x^2 -2x-3)

=> 2x/ (x-3)^2 -1/ (x+1) - 8/ ( x+1)(x-3)

=> [2x(x+1) - (x-3)^2 - 8(x-3)]/(x-3)^2(x+1)

=> (2x^2 + 2x - x^2 - 9 + 6x - 8x +24)/ (x-3)^2(x+1)

=> (x^2 + 15)/ (x-3)^2(x+1)

**Therefore 2x/(x^2 -6x+9) - 1/(x+1) - 8/(x^2 -2x-3) = (x^2 + 15)/ (x-3)^2(x+1).**

We are given the expression :

E = 2x/(x^2 - 6x +9) - 1/(x=1) - 8/(x^2-2x -3)

First let us simplify the denominator (x^2 - 6x +9) = (x-3)^2

Also we will factor the denominator (x^2 -2x+3) = (x-3)(x+1)

Now we will rewrite the expression:

==> E = 2x/(x-3)^2 - 1/(x+1) - 8/(x-3)(x+1)

Now we will determine the common denominator which is (x+1)(x-3)^2

==> E = [2x(x+1) - 1(x-3)^2 - 8(x-3) ] / (x+1)(x-3)^2

= ( 2x^2 + 2x -x + 3 - 8x + 24) / (x+1)(x-3)^2

= (2x^2 -7x + 27) / (x+1)(x-3)^2

** ==> E = (2x^2 - 7x + 27) / (x+1)(x-3)^2 **

To simplify: 2x/(x^2 -6x+9) - 1/(x+1) - 8/(x^2 -2x-3).

First we find the common denominators of x^2-6x+9, (x+1) and (x^2-2x-3).

x^2-6x+9 = (x-3)^2.

So the LCM of x^2-6x+9, (x+1) and (x^2-2x-3) is (x-3)^2(x+1).

Therefore

2x/(x^2-6x+9) = 2x/(x-3)^2 = 2x(x+1)/((x-3)^2(x+1)}...(1)

1/(x+1) = -(x-3)^2/{(x-3)^2(x+1) }...(2)

-8/(x^2-2x+3) = -8/(x-3)(x+1) = -8(x-3)/{(x-3)^2(x+1)}...(3)

(1)+(2)+3): 2x/(x^2 -6x+9) - 1/(x+1) - 8/(x^2 -2x-3) = { 2x(x+1)- (x-3)^2 -8((x-3)}/{(x-3)^2(x+1)}.

RHS numerator = 2x(x+1)-(x-3)^2-8(x-3) = 2x^2+2x-(x^2-6x+9)-8x+24 = 2x^2+2x-x^2+6x-9 - 8x + 24 = (2x^2-x^2)+(2x+6x-8x)-9 +24 = x^2 +15 .

Therefore 2x/(x^2 -6x+9) - 1/(x+1) - 8/(x^2 -2x-3 = (x^2+15)/(x-3)^2(x+1).