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msanyadm
@msanyadm
July 2022
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Не выполняя построения найдите координаты точек пересечения окружности x^2+y^2=5 и прямой х+3у=7
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galina57
Verified answer
x^2+y^2=5 х=7-3у
х+3у=7 (7-3у)^2+y^2=5
49-42y+9y^2+y^2=5
10y^2-42y+44=0
5y^2-21y+22=0
D=441-4*5*22=1
y1=(21-1)/10=2 => x1=7-3*2=1
y2=(21+1)/10=2,2 => x2=7-3*2,2=0,4
Ответ: х1=1; х2=0,4
у1=2; у2=2,2
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Answers & Comments
Verified answer
x^2+y^2=5 х=7-3ух+3у=7 (7-3у)^2+y^2=5
49-42y+9y^2+y^2=5
10y^2-42y+44=0
5y^2-21y+22=0
D=441-4*5*22=1
y1=(21-1)/10=2 => x1=7-3*2=1
y2=(21+1)/10=2,2 => x2=7-3*2,2=0,4
Ответ: х1=1; х2=0,4
у1=2; у2=2,2