<prompt> Acha a,b, c, z ziwe nambari changamano kama vile a b c 0 naaz2 bz c 0.Tafuta thamani kubwa iwezekanavyo ya z.  <response>: Acha r a b c. Tunaweza kupanga upya az2 bz c 0 kamaaz2 -bz - c.Kwa Ukosefu wa Usawa wa Pembetatu,az2 -bz - c le bz c,hivyo az2 le bz c, au rz2 le rz r. Kishaz2 le z 1,hivyo z2 - z - 1 le 0. Sababu hizi kamakushoto z - frac1 - sqrt52 kulia left z - frac1 sqrt52 kulia le 0,hivyo z le frac1 sqrt52.Nambari a 1, b -1, c -1, na z frac1 sqrt52 inakidhi masharti yaliyotolewa, kwa hivyo kubwa zaidi iwezekanavyo. thamani ya z ni frac1 sqrt52. 