Without using a calculator, solve the inequality $\frac{x}{x-2} < \frac{3}{(5x+2)(2-x)}$. Hence, solve the following inequalities.

1. Solve the base inequality from the main stem, as its result is required for the 'Hence' part.

Since the TARGET depends on the result of the main stem, we must first solve $\frac{x}{x-2} < \frac{3}{(5x+2)(2-x)}$.
$\frac{x}{x-2} - \frac{3}{(5x+2)(2-x)} < 0$
$\frac{x}{x-2} + \frac{3}{(5x+2)(x-2)} < 0$
$\frac{x(5x+2) + 3}{(x-2)(5x+2)} < 0$
$\frac{5x^2 + 2x + 3}{(x-2)(5x+2)} < 0$
For the quadratic $5x^2 + 2x + 3$, the discriminant is $2^2 - 4(5)(3) = -56 < 0$. Since the coefficient of $x^2$ is $5 > 0$, the numerator $5x^2 + 2x + 3 > 0$ for all real $x$.
Therefore, the denominator must be negative for the fraction to be less than zero:
$(x-2)(5x+2) < 0$
$-\frac{2}{5} < x < 2$

2. Substitute $x = \text{e}^x$ into the established solution.

By substituting $x$ with $\text{e}^x$ into the solution of the base inequality, we get:
$-\frac{2}{5} < \text{e}^x < 2$

3. Solve the resulting exponential inequality.

Since $\text{e}^x > 0$ for all real $x$, the left inequality $-\frac{2}{5} < \text{e}^x$ is strictly true and imposes no restrictions.
We therefore only need to satisfy the right inequality:
$\text{e}^x < 2$
Taking the natural logarithm on both sides:
$x < \ln 2$

1. Solve the original inequality (required prerequisite for the 'Hence' method).

Let the variable in the original inequality be $u$. We are given:
$\frac{u}{u-2} - \frac{3}{(5u+2)(2-u)} < 0$
$\frac{u(5u+2)+3}{(u-2)(5u+2)} < 0$
$\frac{5u^2+2u+3}{(u-2)(5u+2)} < 0$
Since the discriminant of the numerator is $2^2 - 4(5)(3) = -56 < 0$ and the coefficient of $u^2$ is positive, $5u^2+2u+3 > 0$ for all real $u$.
Thus, the inequality simplifies to the denominator being negative:
$(u-2)(5u+2) < 0$
Which yields the solution:
$-\frac{2}{5} < u < 2$

2. Relate the target inequality to the original inequality by substitution.

Let $u = \frac{x}{2}$. Substituting this into the original expression gives:
$\text{LHS} = \frac{x/2}{x/2-2} = \frac{x}{x-4}$
$\text{RHS} = \frac{3}{(5(x/2)+2)(2-x/2)} = \frac{12}{(5x+4)(4-x)}$
Notice that the target inequality $\frac{x}{x-4} > \frac{12}{(5x+4)(4-x)}$ is identical to the original inequality evaluated at $u = \frac{x}{2}$, but with the inequality sign reversed (from $<$ to $>$).

3. Determine the solution set by applying the substitution and the reversed condition.

Since the original expression was $< 0$ for $-\frac{2}{5} < u < 2$, the reversed condition ($> 0$) holds for the outer regions:
$u < -\frac{2}{5} \quad \text{or} \quad u > 2$
Substitute $u = \frac{x}{2}$ back into the inequalities:
$\frac{x}{2} < -\frac{2}{5} \quad \text{or} \quad \frac{x}{2} > 2$
$x < -\frac{4}{5} \quad \text{or} \quad x > 4$