Problem K. 329. (February 2012)
K. 329. Given that x is a positive real number such that , determine the value of without finding the value of x.
(6 pont)
Deadline expired on March 12, 2012.
Sorry, the solution is available only in Hungarian. Google translation
Megoldás. Használjuk \(\displaystyle \left( x + \frac {1}{x} \right)\) hatványait: \(\displaystyle \left( x + \frac {1}{x} \right)^2=x^2 + \frac {1}{x^2} + 2 =7+2=9\), azaz (\(\displaystyle x\) pozitív!) \(\displaystyle x + \frac {1}{x}=3\). \(\displaystyle \left( x + \frac {1}{x}\right)^5=x^5+5x^3+10x+\frac{10}x +\frac{5}{x^3}+\frac{1}{x^5}\), illetve \(\displaystyle \left( x + \frac {1}{x} \right)^3=x^3+3x+\frac{3}{x} +\frac{1}{x^3}\). A harmadik hatványban felhasználva korábbi eredményünket \(\displaystyle 3^3=x^3+\frac{1}{x^3}+3\cdot 3\), ahonnan \(\displaystyle x^3+\frac{1}{x^3}=27-9=18\). Ezt felhasználva nézzük az ötödik hatványt: \(\displaystyle 3^5=x^5+\frac{1}{x^5}+5\cdot 18 +10 \cdot 3\), azaz \(\displaystyle x^5+\frac{1}{x^5}=243-90-30=123\).
Statistics:
118 students sent a solution. 6 points: 67 students. 5 points: 27 students. 4 points: 1 student. 3 points: 2 students. 2 points: 2 students. 1 point: 5 students. 0 point: 13 students. Unfair, not evaluated: 1 solutions.
Problems in Mathematics of KöMaL, February 2012