7. How many solutions does the equation

\[
\frac{1}{(x-1)^{2}}+\frac{1}{(x-2)^{2}}=\frac{2}{x^{2}} ?
\]

![](https://cdn.mathpix.com/cropped/2024_05_06_948de1dbe9747fa89f7eg-2.jpg?height=446&width=645&top_left_y=131&top_left_x=1322)
Answer: 1.

Solution. Let's introduce the functions $f(x)=\frac{1}{(x-1)^{2}}+\frac{1}{(x-2)^{2}}$ and $g(x)=\frac{2}{x^{2}}$.

For $x<0$, both functions $f$ and $g$ are positive, but $f(x)$ increases from $+\infty$ to 2 (not inclusive), since $f^{\prime}(x)=\frac{-2}{(x-1)^{3}}-\frac{2}{(x-2)^{3}}<0 \quad \text { for } \quad x \in(-\infty ; 0)$, while $g(x)$ decreases from $+\infty$ to 2 (not inclusive), since $g^{\prime}(x)=\frac{-4}{x^{3}}<0 \quad \text { for } \quad x \in(-\infty ; 0)$. Therefore, for $x<0$, the equation $f(x)=g(x)$ has no solutions.

For $0<x<1$, the function $f$ decreases from $+\infty$ to 2 (not inclusive), since $f^{\prime}(x)=\frac{-2}{(x-1)^{3}}-\frac{2}{(x-2)^{3}}<0 \quad \text { for } \quad x \in(0 ; 1)$, while the function $g$ decreases from $+\infty$ to 2 (not inclusive), since $g^{\prime}(x)=\frac{-4}{x^{3}}<0 \quad \text { for } \quad x \in(0 ; 1)$. However, for $0<x<1$, the inequality $f(x)>g(x)$ holds, because $\frac{1}{(x-1)^{2}}>\frac{1}{x^{2}}$ and $\frac{1}{(x-2)^{2}}>\frac{1}{x^{2}}$, so $f(x)>g(x)$. Therefore, for $0<x<1$, the equation has no solutions.

For $x>2$, the function $f$ increases from 2 (not inclusive) to $+\infty$, since $f^{\prime}(x)=\frac{-2}{(x-1)^{3}}-\frac{2}{(x-2)^{3}}>0 \quad \text { for } \quad x \in(2 ; +\infty)$, while the function $g$ decreases from 2 (not inclusive) to 0, since $g^{\prime}(x)=\frac{-4}{x^{3}}<0 \quad \text { for } \quad x \in(2 ; +\infty)$. 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 $f(x)<g(x)$. Therefore, for $x>2$, the equation has no solutions.

## 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 $f^{\prime}(x)=18 x^{2}-42 x+24$, $x=1$ 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 $(0 ; 1)$.