Add.
-$$\frac { x } { x ^ { 2 } - 1 } + \frac { 6 x } { x - x ^ { 2 } }$$
A) $\frac { - 5 x ^ { 2 } - 6 x } { x ( x - 1 ) ( x + 1 ) }$
B) $\frac { - 5 x + 6 } { ( x - 1 ) ( x + 1 ) }$
C) $\frac { - 5 x - 6 } { ( x - 1 ) ( x + 1 ) }$
D) $\frac { - 5 x ^ { 3 } + x ^ { 2 } - 6 x } { \left( x ^ { 2 } - 1 \right) \left( x - x ^ { 2 } \right) }$

Question

Quizplus · Accepted Answer

To add the two fractions, we need to find a common denominator. Factoring the denominators, we have:

$$\frac{x}{x^2-1}+\frac{6x}{x-x^2}=\frac{x}{(x+1)(x-1)}+\frac{6x}{x(1-x)}$$

Since $(x+1)(x-1)$ is already a factor in the first denominator, we only need to add the factor of $x$ that is missing. This gives us:

$$\frac{x}{(x+1)(x-1)}+\frac{6x}{x(1-x)}=\frac{x}{(x+1)(x-1)}-\frac{6x}{x(1-x)}\cdot\frac{-1}{-1}$$

Simplifying the second fraction, we get:

$$\frac{x}{(x+1)(x-1)}-\frac{6x}{x(1-x)}\cdot\frac{-1}{-1}=\frac{x}{(x+1)(x-1)}+\frac{6x}{(x+1)(x-1)}$$

Combining the two fractions, we have:

$$\frac{x}{(x+1)(x-1)}+\frac{6x}{(x+1)(x-1)}=\frac{x+6x}{(x+1)(x-1)}=\frac{-5x}{(x+1)(x-1)}$$

Simplifying, we get the fraction in choice C.

Add - xx2−1+6xx−x2\frac { x } { x ^ { 2 } - 1 } + \frac { 6 x } { x - x ^ { 2 } }x2−1x​+x−x26x​

Add
- $\frac { x } { x ^ { 2 } - 1 } + \frac { 6 x } { x - x ^ { 2 } }$