Постройте график уравнения:
а) $x^2 - 6xy + 8y^2 = 0$;
б) $2x^2 + 5xy + 2y^2 = 0$;
в) $x^2 + xy - 2y^2 = 0$;
г) $3x^2 - 10xy + 3y^2 = 0$.
$x^2 - 6xy + 8y^2 = 0$
$(x^2 - 6xy + 9y^2) - y^2 = 0$
$(x - 3y)^2 - y^2 = 0$
(x − 3y − y)(x − 3y + y) = 0
(x − 4y)(x − 2y) = 0
x − 4y = 0
4y = x
$y = \frac{1}{4}x$
или
x − 2y = 0
2y = x
$y = \frac{1}{2}x$
$2x^2 + 5xy + 2y^2 = 0$
$2x^2 + 4xy + xy + 2y^2 = 0$
$(2x^2 + 4xy) + (xy + 2y^2) = $
$2x(x + 2y) + y(x + 2y) = 0$
(x + 2y)(2x + y) = 0
x + 2y = 0
2y = −x
$y = -\frac{1}{2}x$
2x + y = 0
y = −2x
$x^2 + xy - 2y^2 = 0$
$x^2 + xy - y^2 - y^2 = 0$
$(x^2 - y^2) + (xy - y^2) = 0$
(x − y)(x + y) + y(x − y) = 0
(x − y)(x + y + y) = 0
(x − y)(x + 2y) = 0
x − y = 0
y = x
или
x + 2y = 0
2y = −x
$y = -\frac{1}{2}x$
$3x^2 - 10xy + 3y^2 = 0$
$3x^2 + 6x^2 - 10xy - 2xy + 3y^2 + y^2 - 6x^2 + 2xy - y^2 = 0$
$(9x^2 - 12xy + 4y^2) - (6x^2 - 2xy) - y^2 = 0$
$(3x - 2y)^2 - 2x(3x - y) - y^2 = 0$
$((3x - 2y)^2 - y^2) - 2x(3x - y) = 0$
(3x − 2y − y)(3x − 2y + y) − 2x(3x − y) = 0
(3x − 3y)(3x − y) − 2x(3x − y) = 0
(3x − y)(3x − 3y − 2x) = 0
(3x − y)(x − 3y) = 0
3x − y = 0
y = 3x
или
x − 3y = 0
3y = x
$y = \frac{x}{3}$
Пожауйста, оцените решение