Question:

Step 1: Define the integer and fractional parts of x. Let $[x]$ denote the greatest integer less than or equal to $x$, and $\{x\}$ denote the fractional part of $x$. We know that any real number $x$ can be uniquely expressed as $x = [x] + \{x\}$. Let $[x] = n$, where $n$ is an integer. Then, we have $x = n + \{x\}$. By definition, the fractional part satisfies the inequality $0 \le \{x\} < 1$. Step 2: Substitute these definitions into the given equation. The given equation is $(x-2) [x] = \{x\} - 1$. Substitute $x = n + \{x\}$ and $[x] = n$ into the equation: $$(n + \{x\} - 2) n = \{x\} - 1$$ Step 3: Expand and rearrange the equation. Expand the left side of the equation: $$n^2 + n\{x\} - 2n = \{x\} - 1$$ To solve for $\{x\}$, rearrange the terms by moving all terms containing $\{x\}$ to one side and the rest to the other side: $$n^2 - 2n + 1 = \{x\} - n\{x\}$$ Factor the left side as a perfect square and factor out $\{x\}$ from the right side: $$(n-1)^2 = \{x\}(1-n)$$ Step 4: Analyze the equation based on the value of $n$. We consider two cases based on the value of the term $(1-n)$. **Case 1: $1-n \ne 0 \implies n \ne 1$** If $n \ne 1$, we can divide both sides of the equation $(n-1)^2 = \{x\}(1-n)$ by $(1-n)$: $$\{x\} = \frac{(n-1)^2}{1-n}$$ Since $(n-1)^2 = (-(1-n))^2 = (1-n)^2$, the expression simplifies to: $$\{x\} = \frac{(1-n)^2}{1-n} = 1-n$$ Now, we apply the fundamental property of the fractional part, $0 \le \{x\} < 1$: $$0 \le 1-n < 1$$ This compound inequality can be split into two separate inequalities: 1. $1-n \ge 0 \implies 1 \ge n \implies n \le 1$ 2. $1-n < 1 \implies -n < 0 \implies n > 0$ Combining these two conditions, we get $0 < n \le 1$. Since $n$ must be an integer, and we assumed $n \ne 1$, there are no integer values for $n$ that satisfy $0 < n < 1$. Therefore, there are no solutions for $x$ in this case. **Case 2: $1-n = 0 \implies n=1$** Substitute $n=1$ into the rearranged equation $(n-1)^2 = \{x\}(1-n)$: $$(1-1)^2 = \{x\}(1-1)$$ $$0 = \{x\} \cdot 0$$ This equation is an identity, meaning it is true for any value of $\{x\}$. This implies that if $n=1$, any valid fractional part $\{x\}$ would satisfy this simplified equation. Now, we must also consider the initial condition $[x]=n=1$. If $[x]=1$, then by definition $x$ must be in the interval $1 \le x < 2$. For any $x$ in this interval, its fractional part is $\{x\} = x - [x] = x - 1$. Let's substitute $[x]=1$ and $\{x\}=x-1$ into the original equation $(x-2)[x] = \{x\}-1$: $$(x-2)(1) = (x-1)-1$$ $$x-2 = x-2$$ This is an identity. This means that the original equation is satisfied for all real numbers $x$ for which $[x]=1$. The condition $[x]=1$ implies that $x$ belongs to the interval $[1, 2)$. Step 5: Determine the number of real solutions. From Case 1 ($n \ne 1$), we found no solutions. From Case 2 ($n=1$), we found that all real numbers $x$ in the interval $[1, 2)$ satisfy the equation. The interval $[1, 2)$ contains infinitely many real numbers. Step 6: State the final answer. The number of real values of $x$ satisfying the equation $(x-2) [x] = \{x\} - 1$ is infinite. The final answer is $\boxed{\text{infinite}}$.