how do u solve this x/(x^2-1)+2/(x+1)=1+1/(2x-2) and also this one this is another peoblem (x+2)/(x-5)+6?

Answer 1

For the first one, multiply through on both sides by (x^2 - 1). A bunch of stuff will cancel because (x^2 - 1) = (x + 1)(x - 1) and 2x+2 = 2(x + 1). I think you can get it from there.
For the second one, you should multiply by (x - 5). Good luck!

Answer 2

Q1. x-2/x-5 +6

Taking LCM

(x-2 + 6(x-5) ) / x-5

» (x-2 + 6x-30) / x-5

» (7x-32)/(x-5) ANSWER.

Q2. x/ (x^2-1) + 2/(x+1) = 1 + 1/(2x-2)

now taking the rule a^2- b^2= (a-b) (a+b)
here x^2-1 = x^2 - 1^2 = (x-1) (x+1)

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

Now taking all the x terms on LHS

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

Taking LCM of the denominators

» [2x + 4 (x-1) – (x+1)] / (2 (x-1) (x+1)) = 1

» [2x + 4x -4 – x – 1] / (2 (x-1) (x+1)) = 1

» (5x-5) / (2 (x-1) (x+1)) = 1

Taking 5 common from the numerator

» (5(x-1)) / (2 (x-1) (x+1)) = 1

Cancelling (x-1) from the numerator and denominator

» 5/2(x+1) = 1

Transferring the denominator term on the LHS to RHS and multiplying it with 1

» 5 = 2(x+1)

» 5 = 2x+2

» 5-2 = 2x

» 3 = 2x
Or
2x=3
» x=3/2 = 1.5

Answer 3

Do your own homework.