Чтобы найти нули функции нужно приравнять y к 0 т. е. y=0
<span>(2-x)(2x+3)=0
4x+6-2x^2-3x=0
-2x^2+x+6=0
D=1^2-4×(-2)×6=1+48=49
x1=-1+7/2×(-2)=6/-4=-3/2
x2=-1-7/2×(-2)=-8/-4=2
Нули функции: -3/2; 2
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X^2 / XY + 5Y^2 / XY = - 6XY
X^2 + 5Y^2 = - 6XY
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2X + 7Y = 6
2X = 6 - 7Y
X = 3 - 3,5Y
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( 3 - 3,5Y)^2 + 5Y^2 = - 6Y*( 3 - 3,5Y)
9 - 21Y + 12,25Y^2 + 5Y^2 = - 18Y + 21Y^2
9 - 21Y + 17,25Y^2 = 21Y^2 - 18Y
21Y^2 - 17,25Y^2 - 18Y + 21Y - 9 = 0
3,75Y^2 + 3Y - 9 = 0
D = 9 + 135 = 144 ; V D = 12
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Y1 = ( - 3 + 12 ) : 7,5 = 1,2
Y2 = ( - 15) : 7,5 = ( - 2 )
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X = 3 - 3,5Y
X1 = 3 - 4,2 = - 1,2
X2 = 3 + 7 = 10
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