Problem C. 1480. (April 2018)
C. 1480. Solve the equation
\(\displaystyle \frac{x^3-7x+6}{x-2}=\frac{2x+14}{x+2} \)
on the set of integers.
(5 pont)
Deadline expired on May 10, 2018.
Sorry, the solution is available only in Hungarian. Google translation
Megoldás. A nevezők miatt \(\displaystyle x≠2\) és \(\displaystyle x≠-2\). Alakítsuk szorzattá a harmadfokú kifejezést:
\(\displaystyle x^3-7x+6=x^3-2x^2+2x^2-4x-3x+6=(x-2)(x^2+2x-3)= (x-2)(x+3)(x-1).\)
Így az egyenletünk
\(\displaystyle \frac{(x-2)(x+3)(x-1)}{x-2}=\frac{2x+4+10}{x+2}.\)
Bal oldalon \(\displaystyle (x-2)\neq=0\)-val egyszerűsítve, jobb oldalon \(\displaystyle (x+2)\neq=0\)-val tagonként leosztva:
\(\displaystyle (x+3)(x-1)=2+\frac{10}{x+2}.\)
A bal oldal egész, így a jobb oldal, és ezért \(\displaystyle \frac{10}{x+2}\) is az. Tehát \(\displaystyle z=x+2\) lehetséges értékei \(\displaystyle 10\) osztói: \(\displaystyle -10\), \(\displaystyle -5\), \(\displaystyle -2\), \(\displaystyle -1\), \(\displaystyle 1\), \(\displaystyle 2\), \(\displaystyle 5\), \(\displaystyle 10\).
\(\displaystyle (z+1)(z-3)=2+\frac{10}{z}\)
\(\displaystyle z\) | \(\displaystyle -10\) | \(\displaystyle -5\) | \(\displaystyle -2\) | \(\displaystyle -1\) | \(\displaystyle 1\) | \(\displaystyle 2\) | \(\displaystyle 5\) | \(\displaystyle 10\) |
bal oldal | \(\displaystyle 117\) | \(\displaystyle 32\) | \(\displaystyle 5\) | \(\displaystyle 0\) | \(\displaystyle -4\) | \(\displaystyle -3\) | \(\displaystyle 12\) | \(\displaystyle 77\) |
jobb oldal | \(\displaystyle 1\) | \(\displaystyle 0\) | \(\displaystyle -3\) | \(\displaystyle -8\) | \(\displaystyle 12\) | \(\displaystyle 7\) | \(\displaystyle 4\) | \(\displaystyle 3\) |
Az egyenletnek nincs megoldása az egész számok halmazán.
Statistics:
134 students sent a solution. 5 points: 59 students. 4 points: 22 students. 3 points: 20 students. 2 points: 8 students. 1 point: 13 students. 0 point: 10 students. Unfair, not evaluated: 2 solutionss.
Problems in Mathematics of KöMaL, April 2018