Question

Докажите неравенство x²-3x+y²+3›0Помогите СРОЧНО

Answer 1

Значит левая часть неравенства всегда больше правой части неравенства. Следовательно, исходное неравенство тоже верно, т.к. мы совершали тождественные преобразования.
Что и требовалось доказать

Answer 2

Solving the Inequality x²-3x+y²+3 > 0

To solve the inequality x²-3x+y²+3 > 0, we can use the given search results to guide us in understanding the nature of the inequality and finding its solution.

The search results provide various mathematical expressions and equations, but none of them directly address the given inequality. Therefore, we will solve the inequality based on standard mathematical principles.

Solution:

To solve the inequality x²-3x+y²+3 > 0, we can start by rearranging the terms:

x² - 3x + y² + 3 > 0

We can also rewrite this as:

(x² - 3x) + (y² + 3) > 0

Now, let's complete the square for the x and y terms separately:

For x: 1. We complete the square for the x terms by adding and subtracting (3/2)² inside the parentheses: x² - 3x + (3/2)² - (3/2)² + y² + 3 > 0 (x - 3/2)² - (9/4) + y² + 3 > 0 (x - 3/2)² + y² - (9/4) + 3 > 0 (x - 3/2)² + y² - (9/4) + 12/4 > 0 (x - 3/2)² + y² + 3/4 > 0

For y: 2. Similarly, we complete the square for the y terms by adding and subtracting (3/2)² inside the parentheses: (x - 3/2)² + (y² + 3/4) > 0

Now, we have the inequality in the form of a sum of squares, which is always non-negative. Therefore, the inequality (x²-3x+y²+3) > 0 holds for all real values of x and y.

Conclusion:

Based on the mathematical analysis, the inequality x²-3x+y²+3 > 0 is true for all real values of x and y.