Question

4x/x²-1 - x+1/x-1
please answer quick ​

Answer 1

Answer:

the answer is (-x^2+2x+5)/(x^2+1)

Step-by-step explanation:

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

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

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

{4-(x+1)(x+1)}/(x+1)(x-1)

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

(-x^2+2x+5)/(x^2-1)

Answer 2

Given: $\frac{4x}{x^{2} -1} -\frac{x+1}{x-1}$

To find: the simple form of expression.

Step-by-step explanation:

Step 1 of 4

The denominator of the first term can be represented by using the identity:

$a^{2}-b^{2} =(a+b)(a-b)$

Here a = x and b = 1.

$\frac{4x}{x^{2} -1} -\frac{x+1}{x-1}=\frac{4x}{(x+1)(x-1)} -\frac{x+1}{x-1}$

Step 2 of 4

Now take LCM and (x-1) is common in both the terms.

$\frac{4x}{(x+1)(x-1)} -\frac{x+1}{x-1}=\frac{4x-(x+1)(x+1)}{(x+1)(x-1)}$

Step 3 of 4

Now open the brackets and the sign of the second term will change as a negative sign is in front of the bracket.

$\frac{4x-(x+1)(x+1)}{(x+1)(x-1)}=\frac{4x-(x^{2} +2x+1)}{(x+1)(x-1)}\\\\\=\frac{4x-x^{2} -2x-1}{(x+1)(x-1)}$

The like terms will get subtracted.

$\frac{4x-x^{2} -2x-1}{(x+1)(x-1)}=\frac{-x^{2} +2x-1}{(x+1)(x-1)}$

Step 4 of 4

Now take negative sign common and split the middle term.

$\frac{-x^{2} +2x-1}{(x+1)(x-1)}=\frac{-(x^{2} -2x+1)}{(x+1)(x-1)}\\\\=\frac{-(x^{2} -x-x+1)}{(x+1)(x-1)}\\\\=\frac{-[x(x-1)-1(x-1)]}{(x+1)(x-1)}\\\\=\frac{-[(x-1)(x-1)]}{(x+1)(x-1)}\\\\=\frac{-(x-1)}{(x+1)}$

$\frac{4x}{x^{2} -1} -\frac{x+1}{x-1}=\frac{-(x-1)}{(x+1)}$