Taie ultimul exercitiu ca e gresit!
Continuare!
[tex]\frac{k^{2} }{2*3} +\frac{k^{2}}{3*4} +\frac{k^{2}}{2*4} =54=>\\\frac{k^{2} }{6} +\frac{k^{2}}{12} +\frac{k^{2}}{8} =54=>\\4k^{2}+2k^{2}+3k^{2}=54*24=>\\9k^{2}=54*24=9*6*24=>k^{2}=6*24=144=>k=+12 , k=-12\\=>pt- k=12=>x=6,y=4,z=3\\=>pt -k=-12=>x=-6,y=-4,z=-3[/tex]
Rezultatele sunt corecte, poti sa faci verificare. Daca nu ai invatat numere negative pune k=12.
Bafta!
