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KaRiNa48845789
@KaRiNa48845789
July 2022
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РЕШИТЕ СИСТЕМУ x^2+3y^2-4x-5y-8=0 = x-y+1=0
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ару00
Х-у=-1
х2+3у2-4х-5у-8=0
х=-1+у
(-1+у)2+3у2-4(-1+у)-5у-8=0
1-2у +у2+3у2+4-4у-8=0
4у2-11у-3=0
D=121+4*4*3=169=(13)2(квадраты)
у1,у2=11+-13/4*2=3;-0,25.
=> х1=-1+3=2;
х2=-1-0,25=-1,25.
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Answers & Comments
х2+3у2-4х-5у-8=0
х=-1+у
(-1+у)2+3у2-4(-1+у)-5у-8=0
1-2у +у2+3у2+4-4у-8=0
4у2-11у-3=0
D=121+4*4*3=169=(13)2(квадраты)
у1,у2=11+-13/4*2=3;-0,25.
=> х1=-1+3=2;
х2=-1-0,25=-1,25.