We have to prove that (x/(x+1)+1):(1-3x^2/(1-x^2))=(1-x)/(1-2x)

Starting with the left hand side

(x/(x+1)+1):(1-3x^2/(1-x^2))

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

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

=> [(2x+1)/(x+1)]/[(1-4x^2)/(1-x^2)]

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

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

=> [(1-x)/(1-2x)]

which is the right hand side

**This proves that (x/(x+1)+1):(1-3x^2/(1-x^2))=(1-x)/(1-2x)**

We'll manipulate what's inside the 1st pair of brackets:

x/(x+1) + 1

We'll multiply by (x+1) the number 1:

[x + 1*(x+1)]/(x+1) = (x + x + 1)/(x+1) = (2x+1)/(x+1)

We'll manipulate what's inside the 2nd pair of brackets:

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

We'll multiply (2x+1)/(x+1) by (1-x^2)/(1-4x^2)

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

We'll simplify and we'll get:

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

**[x/(x+1)+1]:[1-3x^2/(1-x^2)]=(1-x)/(1-2x)**