2006 AMC 10B Problem 7

1. Clean up the small fraction on the bottom first. Write $1$ as $\frac{x}{x}$ so both parts have the same denominator:
$$1 - \frac{x-1}{x} = \frac{x}{x} - \frac{x-1}{x} = \frac{x - x + 1}{x} = \frac{1}{x}.$$
2. Now the expression is $\sqrt{x \div \frac{1}{x}}$. Dividing by $\frac{1}{x}$ is the same as multiplying by $x$, so the inside is $x \cdot x = x^2$. We have $\sqrt{x^2}$.
3. Be careful here. The square root sign always gives the number that is zero or positive. For example, if $x = -3$, then $\sqrt{(-3)^2} = \sqrt{9} = 3$, not $-3$.
4. So when $x$ is negative, $\sqrt{x^2}$ is the positive number $-x$ (for $x = -3$, that is $-(-3) = 3$).

The answer is $\boxed{\textbf{(A)}\ -x}$.

Simplifying the fraction is routine. The real point of the problem is that $\sqrt{x^2}$ equals the size of $x$ without its sign. The words "when $x < 0$" are there on purpose: whenever a problem tells you the sign of a variable, expect that fact to decide between $x$ and $-x$.

Plug in a number. Try $x = -2$. The bottom is $1 - \frac{-3}{-2} = 1 - \frac{3}{2} = -\frac{1}{2}$. The inside is $-2 \div \left(-\frac12\right) = 4$, and $\sqrt{4} = 2$. Among the choices, only $-x = 2$ matches. (Do not test with $x = -1$: then choices (A) and (C) both equal $1$, and you could not tell them apart.)

Simplify the inner denominator first:
$$1 - \frac{x-1}{x} = \frac{x - (x-1)}{x} = \frac{1}{x}.$$
Dividing $x$ by $\frac{1}{x}$ gives $x \cdot x = x^2$, so the expression is $\sqrt{x^2}$.

The principal square root of $x^2$ is $|x|$, never just $x$. Since the problem specifies $x < 0$, we have $|x| = -x$, which is a positive number.

The answer is $\boxed{\textbf{(A)}\ -x}$.