Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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Problem C. 845. (March 2006)

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


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 pont)

Deadline expired on April 18, 2006.

Sorry, the solution is available only in Hungarian. Google translation

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.


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