456. Decompose the fractions $\frac{2}{x^{2}-1}$ and $\frac{2 r}{x^{2}-1}$ into the sum of two fractions of the form $\frac{A}{x-1}+\frac{B}{x+1}$.
456. 457) We assume

$$
\frac{2}{x^{2}-1}=\frac{A}{x-1}+\frac{B}{x+1}
$$

Bringing to a common denominator, we have:

$$
\begin{gathered}
A x+A+B x-B=2 \\
x(A+B)+A-B=2
\end{gathered}
$$

Since the left and right sides of the equation must be identically equal for any $x$, then

$$
\begin{aligned}
& A+B=0 \\
& A-B=2
\end{aligned}
$$

thus,

$$
2 A=2
$$

and

$$
A=1 ; B=-1
$$

Therefore:

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

2) Let

$$
\frac{2 x}{x^{2}-1}=\frac{A}{x-1}+\frac{B}{x+1}
$$

Then

$$
2 x=A x+A+B x-B
$$

or

$$
2 x=x(A+B)+A-B
$$

$13^{*}$
thus,

$$
\begin{aligned}
& A+B=2 \\
& A-B=0
\end{aligned}
$$

from which

$$
2 A=2
$$

and

$$
A=1 ; B=1
$$

Therefore

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