Find the set of values of $x$ such that $-1<\dfrac{2x+3}{x-1}<1$. Deduce the range of values of $x$ such that $-1<\dfrac{2x+1}{x-2}<1$.

1. Transform the rational inequality into a quadratic inequality by squaring.

Since the inequality is of the form $-1 < \text{expression} < 1$, it is equivalent to $|\text{expression}| < 1$. We can square both sides:
$-1 < \frac{2x+3}{x-1} < 1$
$\left( \frac{2x+3}{x-1} \right)^2 < 1$
$(2x+3)^2 < (x-1)^2$

2. Solve the quadratic inequality to find the solution set for the first part.

$4x^2 + 12x + 9 < x^2 - 2x + 1$
$3x^2 + 14x + 8 < 0$
$(3x+2)(x+4) < 0$

Therefore, the set of values is:
$-4 < x < -\frac{2}{3}$

3. Identify the relationship between the expressions in the first and second inequalities.

Let $\mathrm{f}(x) = \dfrac{2x+3}{x-1}$.
We want to solve $-1 < \mathrm{f}(X) < 1$ where $\mathrm{f}(X) = \dfrac{2x+1}{x-2}$.
Notice that substituting $X = x-1$ into $\mathrm{f}(X)$ gives:
$\mathrm{f}(x-1) = \frac{2(x-1)+3}{(x-1)-1} = \frac{2x-2+3}{x-2} = \frac{2x+1}{x-2}$

4. Use the substitution to deduce the range of values for the second inequality.

From the first part, the solution to $-1 < \mathrm{f}(X) < 1$ is $-4 < X < -\frac{2}{3}$.
Substituting $X = x-1$ into this range:
$-4 < x-1 < -\frac{2}{3}$
$-3 < x < \frac{1}{3}$