6. How many solutions does the equation

$$
\frac{1}{(x-1)^{2}}+\frac{1}{(x-2)^{2}}=\frac{2}{x^{2}} ?
$$
Answer: 1.

Solution. Let us consider the functions

$$
f(x)=\frac{1}{(x-1)^{2}}+\frac{1}{(x-2)^{2}} \quad \text { and } \quad g(x)=\frac{2}{x^{2}}.
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_2cf4ae7b738ff8febec4g-2.jpg?height=443&width=653&top_left_y=721&top_left_x=1318)

For $x < 0$, both functions $f$ and $g$ are positive, and for $x \in (0 ; 1)$, the function $f$ is decreasing from $+\infty$ to 2 (not including), since $f^{\prime}(x) = -\frac{2}{(x-1)^{3}} - \frac{2}{(x-2)^{3}} < 0$ for $x \in (0 ; 1)$, while the function $g$ is decreasing from $+\infty$ to 2 (not including), since $g^{\prime}(x) = \frac{-4}{x^{3}} < 0$ for $x > 0$. Therefore, for $x \in (0 ; 1)$, the inequality $f(x) > g(x)$ holds, and for $x > 2$, the inequalities $0 < \frac{1}{(x-1)^{2}} < \frac{1}{x^{2}}$ and $\frac{1}{(x-2)^{2}} > \frac{1}{x^{2}}$ hold, so $f(x) > g(x)$, meaning that the equation has no solutions for $x > 2$.

## Second Solution.

The original equation, under the conditions $x \neq 0, x \neq 1, x \neq 2$, is equivalent to the equation $6 x^{3} - 21 x^{2} + 24 x - 8 = 0$. Consider the function $f(x) = 6 x^{3} - 21 x^{2} + 24 x - 8$. Since $f^{\prime}(x) = 18 x^{2} - 42 x + 24$, then $x = 1$ is a point of maximum, and $x = \frac{4}{3}$ is a point of minimum. The function $f$ is increasing on the domain $(-\infty, 1) \cup \left(\frac{4}{3} ; +\infty\right)$ and decreasing on the interval $\left(1 ; \frac{4}{3}\right)$. Since $f(0) = -8, f(1) = 1, f\left(\frac{4}{3}\right) = \frac{8}{9}$, the equation $f(x) = 0$ has a unique root, which lies in the interval $(0 ; 1)$.