Question

Х^2+7х-78/х^2-9 больше или ровно 5

Answer 1

$\frac{x^2+7x-78}{x^2-9} \geq 5 \\ \\ \frac{x^2+7x-78-5x^2+45}{x^2-9} \geq 0 \\ \\ \frac{-4x^2+7x-123}{x^2-9} \geq 0 \\ \\ \frac{4x^2-7x+123}{x^2-9} \leq 0\\ \\ D=49-4*4*123\ \textless \ 0$
⇒ числитель >0 при любом х

$\frac{1}{(x-3)(x+3)} \leq 0 \\ \\ ++++(-3)------(3)+++++$

ОТВЕТ x∈(-3;3)

Answer 2

To solve the inequality $\frac{x^2 + 7x - 78}{x^2 - 9} \geq 5$, we need to find the values of $x$ that satisfy the inequality.

First, let's simplify the expression on the left-hand side:

$\frac{x^2 + 7x - 78}{x^2 - 9}$

To simplify this rational expression, we factor the numerator and denominator:

$\frac{(x+13)(x-6)}{(x+3)(x-3)}$

Now, the inequality becomes:

$\frac{(x+13)(x-6)}{(x+3)(x-3)} \geq 5$

To solve this inequality, we'll consider three intervals based on the critical points of the function: $x = -13$, $x = -3$, and $x = 3$.

Interval 1: $x < -13$ Pick $x = -14$ (a value less than -13):

$\frac{(-14 + 13)(-14 - 6)}{(-14 + 3)(-14 - 3)} = \frac{-1 \times -20}{-11 \times -17} = \frac{20}{187}$

Since $\frac{20}{187} < 5$, this interval does not satisfy the inequality.

Interval 2: $-13 < x < -3$ Pick $x = -5$ (a value between -13 and -3):

$\frac{(-5 + 13)(-5 - 6)}{(-5 + 3)(-5 - 3)} = \frac{8 \times -11}{-2 \times -8} = \frac{88}{16} = 5.5$

Since $5.5 \geq 5$, this interval satisfies the inequality.

Interval 3: $x > -3$ Pick $x = 4$ (a value greater than -3):

$\frac{(4 + 13)(4 - 6)}{(4 + 3)(4 - 3)} = \frac{17 \times -2}{7 \times 1} = \frac{-34}{7}$

Since $\frac{-34}{7} < 5$, this interval does not satisfy the inequality.

So, the solution to the inequality is $-13 < x < -3$ or, in interval notation, $x \in (-13, -3)$.