Question

X^2+4xy+13y^2-6y+1<=0​

Answer 1

Ответ:

$x=-\frac{2}{3};\,\,y=\frac{1}{3}$.

Объяснение:

Выделим полный квадрат

(х+2у)²+9у²-6у+1≤0

(х+2у)²+(3у-1)²≤0

Так как в скобках неотрицательные выражения, то неравенство выполняется лишь в случае равенства нулю обеих скобок.

Из второй скобки

3у-1=0

$y=\frac{1}{3}$

Первая скобка тоже должна равняться нулю

х+2у=0

Подставим значение у в эту скобку

$x+2*\frac{1}{3}=0$

$x=-\frac{2}{3}$

Answer 2

The given inequality is:

x^2 + 4xy + 13y^2 - 6y + 1 ≤ 0

To solve this inequality, you can start by considering it as a quadratic equation in terms of x with y treated as a constant:

x^2 + (4y)x + 13y^2 - 6y + 1 ≤ 0

Now, we can use the discriminant (the part inside the square root in the quadratic formula) to determine the nature of the solutions for x:

Discriminant = (4y)^2 - 4(13y^2 - 6y + 1)

Simplify the discriminant:

16y^2 - 4(13y^2 - 6y + 1) = 16y^2 - (52y^2 - 24y + 4)

Now, simplify further:

16y^2 - 52y^2 + 24y - 4 = -36y^2 + 24y - 4

The discriminant is given by:

D = -36y^2 + 24y - 4

Now, let's find the roots of this discriminant equation:

-36y^2 + 24y - 4 = 0

Divide the equation by -4 to simplify:

9y^2 - 6y + 1 = 0

Now, we can use the quadratic formula to find the values of y:

y = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 9, b = -6, and c = 1:

y = (-(-6) ± √((-6)^2 - 4(9)(1))) / (2(9))

y = (6 ± √(36 - 36)) / 18

y = (6 ± √0) / 18

Since the discriminant is zero, there is only one real root for y:

y = 6/18 = 1/3

Now that we have found the value of y, we can substitute it back into the original inequality:

x^2 + 4xy + 13y^2 - 6y + 1 ≤ 0

x^2 + 4x(1/3) + 13(1/3)^2 - 6(1/3) + 1 ≤ 0

x^2 + (4/3)x + (13/9) - (2/3) + 1 ≤ 0

x^2 + (4/3)x + (13/9) - (6/9) + (9/9) ≤ 0

x^2 + (4/3)x + (7/9) ≤ 0

Now, you have a quadratic inequality in terms of x:

x^2 + (4/3)x + (7/9) ≤ 0

You can solve this quadratic inequality by finding its critical points and testing the intervals between them to determine when it is less than or equal to zero. The solution for this inequality will be a range of values for x.