Solve the following equation:

$$
1-|x+1|=\frac{[x]-x}{|x-1|}
$$
Solve the following equation:

$$
1-|x+1|=\frac{[x]-x}{|x-1|}
$$

I. Solution. The denominator of the fraction cannot be 0, so $x \neq 1$. In the following, we distinguish three cases based on how the absolute value signs can be broken down.

If $x>1$, then $|x+1|=x+1,|x-1|=x-1$, and after rearrangement, (1) becomes:

$$
[x]=2 x-x^{2}
$$

Since $x>1$, it follows that $[x] \geq 1$ also holds, so according to (1a), $2 x-x^{2} \geq 1$, from which, after rearrangement, $(x-1)^{2} \leq 0$ follows. This is only true if $x=1$, which we have excluded.

If $-1 \leq x < 1$, then $|x+1|=x+1,|x-1|=1-x$, and (1) becomes:

$$
1-(x+1)=\frac{[x]-x}{1-x}
$$

Rearranging, we get:

$$
[x]=x-1-\frac{x}{1-x}
$$

Since $x < 1$, the right-hand side is always less than $x-1$, so (1b) implies $x-1-\frac{x}{1-x} < x-1$. This is always true, so $[x]$ can be any integer less than $x-1$. The only integer that fits is $[x]=0$, which gives $x=0$.

If $x < -1$, then $|x+1|=-(x+1),|x-1|=1-x$, and (1) becomes:

$$
1-(-x-1)=\frac{[x]-x}{1-x}
$$

Rearranging, we get:

$$
[x]=x-1-\frac{x}{1-x}
$$

Since $x < -1$, the right-hand side is always less than $x-1$, so (1c) implies $x-1-\frac{x}{1-x} < x-1$. This is always true, so $[x]$ can be any integer less than $x-1$. The possible values for $[x]$ are -2 or -3.

If $[x]=-2$, then (1c) gives $x=-2$, which satisfies all conditions. If $[x]=-3$, then $x=-\sqrt{5}$, which is also a solution.

Thus, the equation has three solutions: $-\sqrt{5}, -2$, and 0.

II. Solution. Multiply both sides of (1) by $-|x-1|$, and plot the functions on both sides of the resulting equation:

$$
|x-1|(|x+1|-1)=x-[x]
$$

1. Figure

![](https://cdn.mathpix.com/cropped/2024_05_02_d7cd2a94787c0a1d466fg-1.jpg?height=597&width=677&top_left_y=1382&top_left_x=651)

2. Figure

![](https://cdn.mathpix.com/cropped/2024_05_02_d7cd2a94787c0a1d466fg-1.jpg?height=677&width=795&top_left_y=2006&top_left_x=600)

3. Figure

![](https://cdn.mathpix.com/cropped/2024_05_02_d7cd2a94787c0a1d466fg-2.jpg?height=586&width=566&top_left_y=130&top_left_x=774)

The first factor on the left side is equal to $-x-2$ if $x \leq -1$, and $x$ if $x > -1$. The second factor is equal to $1-x$ if $x \leq 1$, and $x-1$ if $x > 1$ (1st figure). Thus, the value on the left side is respectively (2nd figure):

$$
(x+2)(x-1), \quad x(1-x), \quad \text{or} \quad x(x-1)
$$

depending on whether

$$
x \leq -2, \quad x \leq 1, \quad \text{or} \quad x > 1
$$

The function on the right side is called the fractional part of $x$, and its value is always between 0 and 1, since $[x]$ cannot be greater than $x$ and cannot be less than $(x-1)$ (3rd figure). Therefore, as the value of $x$ increases from $(-\infty)$, the two functions can first be equal between -3 and -2, since $x$ cannot be equal to the value of the right side for $x > 1$, as the value on the left side is always greater than $(x-1)$, and this is greater than the value on the right side.