<prompt> Acha z_1, z_2, z_3 ziwe nambari changamano kiasi kwamba z_1 2,z_2 3, na z_3 4. Pata thamani kubwa iwezekanavyo yaz_1 - z_22 z_1 - z_32 z_2 - z_32.  <response>: Tuna hiyoanzaalignz_1 jumlaz_1 z_12, z_2 jumlaz_2 z_22, z_3 jumlaz_3 z_32.mwishoalignVile vile,anzaalignz_1 - z_22 z_1 - z_32 z_2 - z_32 z_1 - z_2 muhtasariz_1 - z_2 z_1 - z_3jumlaz_1 - z_3 z_2 - z_3jumlaz_2 - z_3 z_1 - z_2 muhtasariz_1 - muhtasariz_2 z_1 - z_3 overlinez_1 - overlinez_3 z_2 - z_3 jumlaz_2 - jumlaz_3 z_1 muhtasariz_1 - z_1 muhtasariz_2 - muhtasariz_1 z_2 z_2 jumlaz_2 z_1 jumlaz_1 - z_1 jumlaz_3 - muhtasariz_1 z_3 z_1 muhtasariz_3 z_2 muhtasariz_3 - z_2 muhtasariz_3 - overlinez_2 z_3 z_2 overlinez_3 2z_12 2z_22 2z_32 - z_1 overlinez_2 overlinez_1 z_2 z_1 overlinez_3 overline z_1 z_3 z_2 jumlaz_3 jumlaz_2 z_3.mwishoalignSasa,anzaalignz_1 z_2 z_32 z_1 z_2 z_3jumlaz_1 z_2 z_3 z_1 z_2 z_3jumlaz_1 jumlaz_2 jumlaz_3 z_1 muhtasariz_1 z_1 jumlaz_2 z_1 jumlaz_3 z_2 jumlaz_1 z_2 jumlaz_2 z_2 jumlaz_3 z_3 muhtasariz_1 z_3 jumlaz_2 z_3 jumlaz_3 z_12 z_22 z_32 z_1 jumlaz_2 jumlaz_1 z_2 z_1 jumlaz_3 juu ya mstariz _1 z_3 z_2 jumlaz_3 jumlaz_2 z_3.endalignKuongeza milinganyo hii miwili, tunapataz_1 - z_22 z_1 - z_32 z_2 - z_32 z_1 z_2 z_32 3z_12 3z_22 3z_32.Kwa hiyo,anzaalignz_1 - z_22 z_1 - z_32 z_2 - z_32 3z_12 3z_22 3z_32 - z_1 z_2 z_32 le 3 cdoti 22 3 cdoti 32 3 cdoti 42 87.endalignIli usawa ufanyike, ni lazima tuwe na z_1 z_2 z_3 0. Bila hasara ya jumla, tunaweza kudhani kuwa z_1 2. Kisha z_2 z_3 -2. Kuchukua conjugate, tunapata muhtasariz_2 ujumlaz_3 -2.Tangu z_2 3, overlinez_2 frac9z_2. Tangu z_3 4, overlinez_3 frac16z_3, hivyofrac9z_2 frac16z_3 -2.Kisha 9z_3 16z_2 -2z_2 z_3. Inabadilisha z_3 -z_2 - 2, tunapata9-z_2 - 2 16z_2 -2z_2 -z_2 - 2.Hii hurahisisha hadi 2z_22 - 3z_2 18 0. Kwa fomula ya roboduara,z_2 frac3 pm 3i sqrt154.Tukichukua z_2 frac3 3i sqrt154, basi z_3 -frac11 3i sqrt154. Mfano huu unaonyesha kwamba usawa unawezekana, kwa hivyo thamani ya juu ni 87.asyukubwa wa kitengo 1 cm;ukanda wa jozi, ztwo, zthree;eneo 2,0;ztwo 34,3sqrt154;zthree -114,-3sqrt154;choraMduara0,0,2,nyekundu;choraMduara0,0,3,kijani;choraMduara0,0,4,bluu;kuchorazone--ztwo--zthree--cycle;nukta z_1, eneo, E;nuktaz_2, ztwo, N;nuktaz_3, zthree, SW;asyMbadala: Ili usawa utokee, lazima tuwe na z_1 z_2 z_3 0. Bila hasara ya jumla, tunaweza kudhani kuwa z_1 2. Kisha z_2 z_3 -2. Acha z_2 x iy ili z_3 -x - 2 - iy, ambapo x na y ni nambari halisi. Tunahitajianzaalign z_22 x2 y2 9 z_32 x 22 y2 16.endalignTukiondoa mlingano wa kwanza kutoka wa pili, tunapata 4x 4 7, au x dfrac34. Suluhisho moja ni z_2 dfrac34 idfrac3sqrt15 4 na z_3 -dfrac114 idfrac3sqrt154. Mfano huu unaonyesha kwamba usawa unawezekana, kwa hivyo thamani ya juu zaidi ni 87. 