B3. The expression is $Z=5 a^{-x}\left(1-a^{-x}\right)^{-1}-3 a^{-x}\left(1+a^{-x}\right)^{-1}-2 a^{x}\left(a^{2 x}-1\right)^{-1}$, where $a^{x} \neq 0,1,-1$.

a) Simplify the expression $Z$.

b) Calculate the value of the expression $Z$ for $a=9^{b+c} \cdot 3^{2 b+c}: 27^{\frac{4}{3} b+c+\frac{1}{3}}$ and $x=1$.

19th Knowledge Competition

in mathematics for students of secondary technical and vocational schools

National Competition, April 13, 2019

## Problems for 3rd Year

Time for solving: 120 minutes. In section A, we will award three points for each correct answer, and deduct one point for each incorrect answer. Write your answers for section A in the left table, leave the right table blank.

![](https://cdn.mathpix.com/cropped/2024_06_07_9fbd0ae098c18b73b6bfg-10.jpg?height=165&width=319&top_left_y=657&top_left_x=640)

| B1 | B2 | B3 |
| :--- | :--- | :--- |
|  |  |  |
B3. We simplify each term separately. The first term is transformed into

$$
5 a^{-x}\left(1-a^{-x}\right)^{-1}=\frac{5}{a^{x}}\left(\frac{a^{x}-1}{a^{x}}\right)^{-1}=\frac{5}{a^{x}-1}
$$

Similarly, the second term is transformed into

$$
3 a^{-x}\left(1+a^{-x}\right)^{-1}=\frac{3}{a^{x}+1}
$$

We factor the denominator of the third term into a product of a sum and a difference

$$
2 a^{x}\left(a^{2 x}-1\right)^{-1}=\frac{2 a^{x}}{a^{2 x}-1}=\frac{2 a^{x}}{\left(a^{x}-1\right)\left(a^{x}+1\right)}
$$

We observe that this is also the common denominator for the first two terms. We extend the first two terms, simplify the expression, and obtain the result

$$
\frac{5}{a^{x}-1}-\frac{3}{a^{x}+1}-\frac{2 a^{x}}{\left(a^{x}-1\right)\left(a^{x}+1\right)}=\frac{8}{\left(a^{x}-1\right)\left(a^{x}+1\right)}=\frac{8}{a^{2 x}-1}
$$

We calculate the value of the expression

$$
a=9^{b+c} \cdot 3^{2 b+c}: 27^{\frac{4}{3} b+c+\frac{1}{3}}=3^{2 b+2 c} \cdot 3^{2 b+c}: 3^{3\left(\frac{4}{3} b+c+\frac{1}{3}\right)}=\frac{1}{3} .
$$

We calculate the value of the expression \( Z \) for the obtained values of \( a \) and \( x: \frac{8}{\left(\frac{1}{3}\right)^{2}-1}=-9 \).
Simplified first term \(\frac{5}{a^{x}-1}\) ..... 1 point
Simplified second term \(\frac{3}{a^{x}+1}\) ..... 1 point
Simplified third term \(\frac{2 a^{x}}{a^{2 x-1}}\) ..... 1 point
Extending all terms to a common denominator ..... 1 point
Transforming the expression into \(\frac{8}{a^{2 x}-1}\) or \(\frac{8}{\left(a^{x}-1\right)\left(a^{x}+1\right)}\) ..... 1 point
Calculated value of the expression \( a=\frac{1}{3} \) ..... \(1^*+1\) points
Calculated value -9 of the expression \( Z \) for \( a=\frac{1}{3} \) and \( x=1 \) ..... 1 point

## 19th Mathematics Knowledge Competition for Students of Secondary Technical and Vocational Schools  National Competition, April 13, 2019

## Solutions and Scoring Guide

(April 5, 2019, 10:30)

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 considered any procedure 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 (it is not possible to arrive at it without calculation), even if it is correct, it is scored with 0 points.

A competitor who has only partially solved the problem, from a generally correct solving procedure but without a clear path to the final solution of the problem, cannot receive more than half of the possible points.

The mark '*' next to the points means that the point or points can be awarded to the competitor for a correct procedure, even if the calculation is incorrect.

## Third Year

![](https://cdn.mathpix.com/cropped/2024_06_07_9fbd0ae098c18b73b6bfg-24.jpg?height=137&width=220&top_left_y=357&top_left_x=929)