Solve the following inequalities:
a) $\frac{x(x+1)}{x-1}>3$;
b) $\frac{x(x+1)}{x-1}>6$
I. Solution: a) Our inequality can also be written as

$$
\frac{x(x+1)}{x-1}-3=\frac{x(x+1)-3(x-1)}{x-1}=\frac{x^{2}-2 x+3}{x-1}=\frac{(x-1)^{2}+2}{x-1}>0
$$

We exclude the value $x=1$, because in this case the fraction on the left side is not defined. Since the numerator is always positive, and the denominator is positive only when $x>1$, our inequality is true

$$
x>1
$$

and only then.

b) We again transform our inequality similarly

$$
\frac{x(x+1)}{x-1}-6=\frac{x^{2}-5 x+6}{x-1}=\frac{(x-2)(x-3)}{x-1}>0
$$

We must again exclude the value $x=1$.

The left side is positive if

1. Both factors of the numerator and the denominator are positive. This is true when

$$
x>3
$$

2. Both factors of the numerator are negative and the denominator is positive. This is the case when

$$
1<x<2
$$

Therefore, the solution is

$$
1<x<2 \quad \text{or} \quad x>3
$$

---

If we consider the inequality in parts:

a) We exclude the value $x=1$.

1. If $x \geq 1$, or $x>1$, then multiplying both sides of our inequality by $(x-1)$, the inequality sign remains unchanged:

$$
x(x+1)>3(x-1)
$$

which simplifies to

$$
x^{2}-2 x+3=(x-1)^{2}+2>0
$$

This is true for all values of $x$.

2. If $x<1$, then multiplying both sides by $(x-1)$, the inequality sign reverses:

$$
x(x+1)<3(x-1)
$$

which simplifies to

$$
x^{2}-2 x+3=(x-1)^{2}+2>0
$$

This is also true for all values of $x$.

Thus, the solution is

$$
x>1
$$

b) We again exclude the value $x=1$.

1. If $x \geq 1$, or $x-1>0$, then

$$
x(x+1)>6(x-1)
$$

which simplifies to

$$
x^{2}-5 x+6=(x-2)(x-3)>0
$$

This is true when

$$
x>3, \quad \text{or} \quad 1<x<2
$$

Therefore, the solution is

$$
1<x<2 \quad \text{or} \quad x>3
$$