Екстремум функції F(x;y)3x^2+5xy+9y^2-6x+2y-7

Ответ:

Пошаговое объяснение:

) (3x - 8y)^2 + 6x * (9x + 8y) = 9x^2 - 48xy + 64y^2 + 54x^2 + 48xy = 63x^2 + 64y^2;

2) (-7x + 2y)^2 - 14x * (3x - 2y) = 49x^2 - 28xy + 4y^2 - 42x^2 + 28xy = 7x^2 + 4y^2;

3) (-5x - y)^2 - 10x * (-7x + y) = 25x^2 + 10xy + y^2 + 70x^2 - 10xy = y^2 + 95x^2;

4) (-8x + 5y)^2 - 16x * (-8x - 5y) = 64x^2 - 80xy + 25y^2 + 128x^2 + 80xy = 192x^2 + 25y^2;

5) y * (7y + 4x) - (2x + y)^2 = 7y^2 + 4xy - 4x^2 - 4xy - y^2 = 6y^2 - 4x^2;

6) y * (7y + 10x) - (-x - 5y)^2 = 7y^2 + 10xy - x^2 - 10xy - 25y^2 = -18y^2 - x^2;

7) y * (3y + 16x) - (-8x - y)^2 = 3y^2 + 16xy - 64x^2 - 16xy - y^2 = 2y^2 - 64x^2;

8) y * (-7y + 6x) - (-3x - y)^2 = -7y^2 + 6xy - 9x^2 - 6xy - y^2 = -8y^2 - 9x^2.