I know how to solve the inequality x > (2/(x-1)) but how can I know not to mutiply all by (x-1)^2? That is a method for solving some inequalities.
you may multiply by $\displaystyle (x-1)^2$ provided you understand $\displaystyle x \ne 1$ and you can deal algebraically with the resulting cubic expression.
In this case, I would prefer to use this method ...
$\displaystyle x > \frac{2}{x-1}$
$\displaystyle x - \frac{2}{x-1} > 0$
$\displaystyle \frac{x(x-1)}{x-1} - \frac{2}{x-1} > 0$
$\displaystyle \frac{x(x-1) - 2}{x-1} > 0$
$\displaystyle \frac{x^2 - x - 2}{x-1} > 0$
$\displaystyle \frac{(x-2)(x+1)}{x-1} > 0$
critical values are $\displaystyle x = 2$ , $\displaystyle x = -1$ , and $\displaystyle x = 0$
these three values are where the expression on the left side of the inequality equals 0 or is undefined.
last step is to check a value from each interval between the critical values to see if the values in that interval make the original inequality true or false.
Suppose $\displaystyle x<1$
Then if you multiply both sides of the inequality by $\displaystyle (x-1)$
you would have to reverse the inequality.
This is because $\displaystyle x-1$ is negative when $\displaystyle x<1$
$\displaystyle 5>4$ but $\displaystyle -5<-4$
The reversal happens if you change the sign of both sides,
which happens when you multiply both sides by a negative value,
or divide both sides by a negative value.
When you are dealing with inequalities, expressions involving x may be positive for certain x
and negative for other x.
Hence, if you multiply both sides by a square, you avoid that scenario (since real squares are non-negative).
Adding and subtracting the same value to both sides does not introduce any sign reversal,
nor does multiplying by 1, as shown by skeeter.