Решить систему x^3-y^3=98 { xy^2-yx^2=-30 - вопрос №339822302013936 от aruukealtymysh 08.02.2023 12:34

Question

Алгебра: Решить систему x^3-y^3=98 { xy^2-yx^2=-30

aruukealtymysh · Answer

X^3-y^3=98xy^2-yx^2=-30(x-y)(x^2+xy+y^2)=98xy(y-x)=-30xy(x-y)=30x-y=30/xy(30/xy)(x^2+xy+y^2)=9830(x ^ 2+xy+y ^ 2)=98xy30x ^2-68xy+30y ^ 2=0       y 030x ^ 2/y ^ 2 - 68xy/y ^ 2+30=0(x/y)=a30a ^ 2-68a+30=015a ^ 2-34a+15=0d=34 ^ 2-4*15*15=256=16 ^ 2a1=5/3a2=3/51) x/y=5/3x=5y/3(5y/3) ^ 3-y ^ 3=98(125y ^ 3-27y ^ 3 )/27=9898y ^3 /27=98y ^3/27=1y ^ 3=27y=3    =    x=52) x/y=3/5x=3y/5(3y/5) ^3-y ^ 3=9827y ^ 3/125 y ^3=98(27y ^ 3-125y ^ 3)/125=98-98y ^ 3/125=98-y ^ 3/125=1y ^ 3=-125y=-5   =   x= -3...