### Instruction: Find all real numbers $x$ such that \[\left(\frac{x}{2x-1}\right)^2 \le 9.\].### Response: I want to solve this inequality involving a quadratic expression in the numerator and a linear expression in the denominator.
To make things easier, I will try to isolate the quadratic expression by taking the square root of both sides.
However, I have to remember that when I take the square root of both sides of an inequality, I need to consider both the positive and negative roots, since they both satisfy the original inequality.
That means I get two inequalities: \[\frac{x}{2x-1} \le 3 \quad \text{or} \quad \frac{x}{2x-1} \ge -3.\]
Now I have two inequalities that look more manageable.
To solve each one, I will try to cross-multiply and simplify.
However, I have to be careful with the signs of the denominator, since they can affect the direction of the inequality.
So I will consider three cases: $x < \frac{1}{2}$, $x = \frac{1}{2}$, and $x > \frac{1}{2}$.
If $x < \frac{1}{2}$, then $2x-1$ is negative, so I have to flip the direction of the inequality when I cross-multiply.
That means the first inequality becomes \[x \ge 6x - 3 \quad \text{or} \quad -5x \ge -3 \quad \text{or} \quad x \le \frac{3}{5}.\]
The second inequality becomes \[x \le -6x + 3 \quad \text{or} \quad 7x \le 3 \quad \text{or} \quad x \le \frac{3}{7}.\]
So in this case, the solution is the intersection of these two inequalities, which is \[x \le \frac{3}{7}.\]
If $x = \frac{1}{2}$, then $2x-1$ is zero, so the fraction is undefined and the inequality does not hold.
So there is no solution in this case.
If $x > \frac{1}{2}$, then $2x-1$ is positive, so I do not have to flip the direction of the inequality when I cross-multiply.
That means the first inequality becomes \[x \le 6x - 3 \quad \text{or} \quad -5x \le -3 \quad \text{or} \quad x \ge \frac{3}{5}.\]
The second inequality becomes \[x \ge -6x + 3 \quad \text{or} \quad 7x \ge 3 \quad \text{or} \quad x \ge \frac{3}{7}.\]
So in this case, the solution is the intersection of these two inequalities, which is \[x \ge \frac{3}{5}.\]
Therefore, the final solution is the union of the two cases where $x \neq \frac{1}{2}$, which is \[x \le \frac{3}{7} \quad \text{or} \quad x \ge \frac{3}{5}.\]