Question

X^2+y^2-10x+4y+30<0 решите пжл​

Answer 1

Ответ:

10х^3+4у^3+3<0

Объяснение:

Возможно неправильно!

Answer 2

To solve the inequality $x^2 + y^2 - 10x + 4y + 30 < 0$, we can follow these steps:

Step 1: Complete the square for both $x$ and $y$ terms. Step 2: Rewrite the inequality in the form $(x - h)^2 + (y - k)^2 < r^2$, where $(h, k)$ is the center and $r$ is the radius of the circle. Step 3: Determine the solution by identifying the region inside the circle.

Let's start with Step 1:

Step 1: Complete the square for $x$ and $y$:

For $x^2 - 10x$, we need to add and subtract $(10/2)^2 = 25$ inside the parentheses: $x^2 - 10x + 25 - 25$.

For $y^2 + 4y$, we need to add and subtract $(4/2)^2 = 4$ inside the parentheses: $y^2 + 4y + 4 - 4$.

Now, rewrite the inequality:

$(x^2 - 10x + 25) + (y^2 + 4y + 4) + 30 - 25 - 4 < 0$.

Simplify:

$(x - 5)^2 + (y + 2)^2 + 5 < 0$.

Now the inequality is in the form $(x - h)^2 + (y - k)^2 < r^2$.

Step 2: The center of the circle is $(h, k) = (5, -2)$, and the square of the radius is $r^2 = 5$.

Step 3: Determine the solution:

Since the square of the radius $r^2$ is positive, the inequality represents an empty circle (no points lie inside the circle). And since we have the strict inequality ($<$), no points on the circle boundary are included in the solution either.

Therefore, the solution to the inequality $x^2 + y^2 - 10x + 4y + 30 < 0$ is an empty set, meaning there are no solutions that satisfy the inequality. The region inside the circle is completely excluded.