Mathematical and Physical Journal
for High Schools
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Problem K. 329. (February 2012)

K. 329. Given that x is a positive real number such that x^2+\frac{1}{x^2}=7, determine the value of x^5+\frac{1}{x^5} 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 \(\displaystyle x^5+\frac{1}{x^5}=243-90-30=123\).


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