Question

82. a) x2 - 6x + 9 > 0;​

Answer 1

Решение:

a) x2 - 6x + 9 > 0

-4х+9>0

-4х>0-9

-4х>-9 | делим обе стороны на (-4), меняем > на <

х<2,25

х принадлежит (-&; 2,25)

Ответ: (-&; 2,25).

Answer 2

To solve the inequality $x^2 - 6x + 9 > 0$, you can follow these steps:

1. First, recognize that the expression $x^2 - 6x + 9$ can be factored as $(x - 3)^2$. This is because it is a perfect square trinomial.

2. Now, rewrite the inequality with the factored expression: $(x - 3)^2 > 0$.

3. Realize that a square of any real number is always non-negative, and it is equal to zero only when the real number itself is zero. Therefore, $(x - 3)^2$ will be greater than zero for all real values of $x$ except when $x - 3 = 0$, which means $x = 3$.

4. So, the solution to the inequality is all real numbers $x$ except $x = 3$. In interval notation, this can be expressed as:

   $-\infty < x < 3$ or $3 < x < \infty$.

In set-builder notation, you can write it as:

$\{x \in \mathbb{R} : x \neq 3\}$.