Question

Determinati numerele naturale a,b,c stiind ca:
$\frac{a ^{3} }{8} = \frac{b ^{2} }{4} = \frac{c}{2}$
si
$\frac{1}{a} + \frac{1}{b} = \frac{24}{c}$

Answer 1

[tex]\frac{b^2}{4}=\frac{c}{2}\Rightarrow b^2=2c,c=\frac{b^2}{2}\\ \frac{a^3}{8}=\frac{c}{2} \Rightarrow a^3=4c\\ \frac{1}{a}+\frac{1}{b}=\frac{24}{c}\Leftrightarrow \frac{1}{a}+\frac{1}{b}=\frac{48}{b^2}\\ \frac{1}{a}=\frac{48}{b^2}-\frac{1}{b}=\frac{48-b}{b^2}\\ b^2=a(48-b)\ |\cdot a^2\\ a^2b^2=a^3(48-b)\\ a^2\cdot2c=4c(48-b)\\ a^2=2(48-b)=96-2b [/tex]
Tinand seama ca a,b,c sunt numere naturale de aici putem face prin incercari(a²<96⇒a≤9).
Obtinem singura varianta convenabila a=8, b=16, c=128, sunt numerele cautate.