Question

Решите неравенство -15/(x+1)^2-3>=0

Answer 1

-15/(х²+1+2х)≥3 |:3

-5/(х²+1+2х)≥1

-5≥х²+1+2х

х²+2х+6≤0

Д=4-4*6=4-26=-22＜0 значит нет корней

старший коэффициент а=1＞0

значит парабола находится выше оси ОХ и не имеет пересечений с ней

ответ: нет решений

Answer 2

To solve the inequality $-\frac{15}{(x+1)^2} - 3 \geq 0$, follow these steps:

Step 1: Move the constant term to the right side of the inequality: $-\frac{15}{(x+1)^2} \geq 3$

Step 2: Get rid of the negative sign by multiplying both sides by -1 (since we are multiplying by a negative number, the inequality sign will flip): $\frac{15}{(x+1)^2} \leq -3$

Step 3: Now, get rid of the fraction by taking the reciprocal of both sides (again, the inequality sign will flip): $(x+1)^2 \geq \frac{1}{-3} = -\frac{1}{3}$

Step 4: Since the square of any real number is non-negative, we can ignore the negative sign and just look at the magnitude: $(x+1)^2 \geq \frac{1}{3}$

Step 5: Take the square root of both sides (don't forget the ± sign): $x + 1 \geq \pm \sqrt{\frac{1}{3}}$

Step 6: Simplify the square root: $x + 1 \geq \pm \frac{\sqrt{3}}{3}$

Now, we need to consider both cases:

Case 1: $x + 1 \geq \frac{\sqrt{3}}{3}$ Subtract 1 from both sides: $x \geq \frac{\sqrt{3}}{3} - 1$

Case 2: $x + 1 \leq -\frac{\sqrt{3}}{3}$ Subtract 1 from both sides: $x \leq -\frac{\sqrt{3}}{3} - 1$

So the solution to the inequality is the combined solution of the two cases:

$-\frac{\sqrt{3}}{3} - 1 \leq x \leq \frac{\sqrt{3}}{3} - 1$

This can also be written as: $x \in \left[ -\frac{\sqrt{3}}{3} - 1, \frac{\sqrt{3}}{3} - 1 \right]$