2. For a real number $x$, where $x \notin\left\{0, \frac{1}{2}, \frac{2}{3}, 1,2\right\}$, simplify the expression

$$
\left(\frac{3 x^{-\frac{1}{3}}}{x^{\frac{2}{3}}-2 x^{-\frac{1}{3}}}-\frac{x^{\frac{1}{3}}}{x^{\frac{4}{3}}-x^{\frac{1}{3}}}\right)^{-1}-\left(\frac{1-2 x}{3 x-2}\right)^{-1}
$$

and calculate its value for $x=\sqrt{3}$. Simplify the final result and rationalize the denominator if necessary.

17th Mathematics Competition for Students of Secondary Technical and Vocational Schools, Selection Competition, March 16, 2017

# Problems for 3rd Year 

Time for solving: 45 minutes.
2. In the denominator of the fraction, we factor out the power of $\mathrm{z}$ with the smallest exponent $\left(\frac{3 x^{-\frac{1}{3}}}{x^{-\frac{1}{3}}(x-2)}-\frac{x^{\frac{1}{3}}}{x^{\frac{1}{3}}(x-1)}\right)^{-1}-\left(\frac{1-2 x}{3 x-2}\right)^{-1}$, simplify the fractions, expand the fraction in the parentheses to a common denominator, and write the reciprocal of the fractions $\left(\frac{(x-2)(x-1)}{2 x-1}\right)-\left(\frac{3 x-2}{1-2 x}\right)$. Then we determine the common denominator again and simplify the expression. This gives us the fraction $\frac{x^{2}}{2 x-1}$. We then substitute $x=\sqrt{3}$ and rationalize the denominator $\frac{3}{2 \sqrt{3}-1} \cdot \frac{2 \sqrt{3}+1}{2 \sqrt{3}+1}$. We get the solution $\frac{6 \sqrt{3}+3}{11}$.

## Factoring out the power of $z$ with the smallest exponent in each of the first two

 fractions $1+1$ points $\qquad$![](https://cdn.mathpix.com/cropped/2024_06_07_115a7be78276e1d4adcag-13.jpg?height=52&width=1725&top_left_y=2578&top_left_x=177)
Considering the reciprocal of the fractions $\left(\frac{1-2 x}{3 x-2}\right)^{-1}=\frac{3 x-2}{1-2 x}$ ..... 1 point
Expanding the fractions to a common denominator $2 x-1$ ..... 1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_115a7be78276e1d4adcag-14.jpg?height=86&width=1756&top_left_y=217&top_left_x=161)

![](https://cdn.mathpix.com/cropped/2024_06_07_115a7be78276e1d4adcag-14.jpg?height=60&width=1753&top_left_y=284&top_left_x=160)

Multiplying the fractions ....................................................................................................................

![](https://cdn.mathpix.com/cropped/2024_06_07_115a7be78276e1d4adcag-14.jpg?height=69&width=1756&top_left_y=385&top_left_x=161)

Note 1: The problem can also be solved by converting powers with rational exponents to roots.

Note 2: Instead of factoring out the power of $\mathrm{z}$ with the smallest exponent, we could have obtained a fraction for the negative exponent.

2nd method.

$$
\begin{aligned}
& \left(\frac{3 x^{-\frac{1}{3}}}{x^{\frac{2}{3}}-2 x^{-\frac{1}{3}}}-\frac{x^{\frac{1}{3}}}{x^{\frac{4}{3}}-x^{\frac{1}{3}}}\right)^{-1}-\left(\frac{1-2 x}{3 x-2}\right)^{-1}=\left(\frac{\frac{3}{\sqrt[3]{x}}}{(\sqrt[3]{x})^{2}-\frac{2}{\sqrt[3]{x}}}-\frac{\sqrt[3]{x}}{x \cdot \sqrt[3]{x}-\sqrt[3]{x}}\right)^{-1}-\frac{3 x-2}{1-2 x}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_06_07_115a7be78276e1d4adcag-14.jpg?height=203&width=779&top_left_y=989&top_left_x=1004)

$$
\begin{aligned}
& =\left(\frac{3}{x-2}-\frac{1}{x-1}\right)^{-1}-\frac{3 x-2}{1-2 x} \\
& =\left(\frac{2 x-1}{(x-2)(x-1)}\right)^{-1}-\frac{3 x-2}{1-2 x} \\
& =\frac{(x-2)(x-1)}{2 x-1}+\frac{3 x-2}{2 x-1}=\frac{x^{2}}{2 x-1} .
\end{aligned}
$$

Writing the power as a root $x^{\frac{1}{3}}=\sqrt[3]{x}$ ..... 1 point
Considering the reciprocal of the fractions $\left(\frac{1-2 x}{3 x-2}\right)^{-1}=\frac{3 x-2}{1-2 x}$ ..... 1 point
Expanding to a common denominator $\frac{(\sqrt[3]{x})^{3}-2}{\sqrt[3]{x}}$ ..... 1 point
Partial root extraction $\sqrt[3]{x^{4}}=x \cdot \sqrt[3]{x}$ ..... 1 point
Simplifying the fractions ..... 1 point
Transforming the first term to the form $\left(\frac{2 x-1}{(x-2)(x-1)}\right)^{-1}$ ..... 1 point
Calculating the fraction $\frac{x^{2}}{2 x-1}$ ..... 1 point
Rationalizing the denominator after substituting $x=\sqrt{3}$ ..... 1 point
Multiplying the fractions ..... $1^{*}$ point
Solution $\frac{6 \sqrt{3}+3}{11}$ ..... 1 point

## 17th Mathematics Knowledge Competition for Students of Secondary Technical and Vocational Schools

Selection Competition, March 16, 2017

## Solutions and Scoring Guide

(16. MARCH 2017, 00:04)

A competitor who arrives at the solution by any correct method (even if the scoring guide does not provide for it) receives all possible points.

A correct method is one that:

- sensibly takes into account the wording of the problem,
- leads to the solution of the problem,
- is mathematically correct and complete.

If an intermediate or final result can be recognized, guessed, read from a diagram, or calculated mentally, the competitor generally receives all the points provided. If, however, the solution is guessed (cannot be arrived at without calculation), and there is no written verification or proof, it is scored with 0 points.

A competitor who has only partially solved the problem, from correct but incomplete solution steps, cannot receive more than half of the possible points.

The mark ${ }^{\prime * \prime}$ next to the points means that the competitor can earn the point or points for a correct procedure, even if the calculation is incorrect.

## Third Year