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July 2022
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решить систему {x^2 + x y + y^2 = 91, x + sqrt(x y) + y = 13}
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skyne8
Verified answer
{x^2+2xy+y^2 -xy=91,
x+√xy+y=13},
{(x+y)^2 -xy=91, (x+y)+√(xy)=13},
Делаем замену: x+y=p, √(xy)=q,
{p^2 -q^2=91, p+q=13},
{(p-q)(p+q)=91, p+q=13},
{(p-q)13=91, p+q=13},
{p-q=7, p+q=13}, {p=10, q=3},
{x+y=10, √(xy)=3},{x+y=10,xy=9},
{x=10-y,(10-y)y=9},{x=10-y,y^2-10y+9=0}
{x=10-y, (y-9)(y-1)=0},{x=10-y,y=9,y=1},
{x=1,y=9} или {x=9,y=1}
5 votes
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oganesbagoyan
можно и без замены {(x+y)² -xy=91, (x+y)+√(xy)=13} ⇔{(x+y+√xy)(x+y-√(xy)=91, x+y +√(xy) =13}
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Answers & Comments
Verified answer
{x^2+2xy+y^2 -xy=91,x+√xy+y=13},
{(x+y)^2 -xy=91, (x+y)+√(xy)=13},
Делаем замену: x+y=p, √(xy)=q,
{p^2 -q^2=91, p+q=13},
{(p-q)(p+q)=91, p+q=13},
{(p-q)13=91, p+q=13},
{p-q=7, p+q=13}, {p=10, q=3},
{x+y=10, √(xy)=3},{x+y=10,xy=9},
{x=10-y,(10-y)y=9},{x=10-y,y^2-10y+9=0}
{x=10-y, (y-9)(y-1)=0},{x=10-y,y=9,y=1},
{x=1,y=9} или {x=9,y=1}