Problem C. 845.

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C. 845. In an algebra class on the first of April, the students were practising. The task was to simplify the fraction

$\frac{{(x+2)}^3+{(y+x)}^3}{{(x+2)}^3-{(y-2)}^3}.$

Agnes, who was the best at mathematics in the class, suggested that if the denominator is not zero, they should just cross out the threes in all the indices, that is, write

$\frac{x+2+y+x}{x+2-(y-2)}= \frac{2x+y+2}{x-y+4}.$

Check the result.

(5 points)

Deadline expired.

Sorry, the solution is published in Hungarian only.

Megoldás: Természetesen ez az ,,egyszerűsítés" nem megengedhető, hibás lépés. Végezzünk átalakításokat (az a³+b³ és az a³-b³ szorzattá bontását alkalmazzuk):

${(x+2)^3+(y+x)^3\over (x+2)^3-(y-2)^3}={(2x+y+2)((x^2+2x+4-2y+xy+y^2)\over(x-y+4)(x^2+2x+4-2y+xy+y^2)}={2x+y+2\over x-y+4}.$

Vagyis a végeredmény jó.

Statistics on problem C. 845.

292 students sent a solution.

5 points: 264 students.

4 points: 5 students.

3 points: 4 students.

2 points: 3 students.

1 point: 5 students.

0 point: 11 students.

Problems in Mathematics of KöMaL, March 2006

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