Question

(x^2+2x)(2x+2)-9*(2x+2)/(x^2-2)≥0

Answer 1

x^2-9=(x-3)*(x+3). x не должен равняться (-3) и 3 (при этих значениях знаменатель обращается в 0 и выражение не имеет смысла). получаем: x^2=12-x; x^2+x-12=0; D=1^2-4*1*(-12)=1+48=49; x1=(-1-7)/2, x2=(-1+7)/2. x1= -4, x2=3. Ответ: x= -4.

Answer 2

To solve the inequality $(x^2+2x)(2x+2) - \frac{9(2x+2)}{x^2-2} \geq 0$, we need to find the values of $x$ that satisfy the inequality. To do that, let's follow these steps:

Step 1: Factorize the expressions. Step 2: Determine the domain of the inequality. Step 3: Find the critical points (points where the expression is equal to zero or undefined). Step 4: Test intervals and determine the sign of the expression. Step 5: State the solution to the inequality.

Let's start with step 1:

Step 1: Factorize the expressions. Factorizing the first expression: $(x^2+2x)(2x+2) = 2x(x^2+2x+1) = 2x(x+1)^2$

Factorizing the second expression: $\frac{9(2x+2)}{x^2-2} = \frac{18(x+1)}{(x-\sqrt{2})(x+\sqrt{2})}$

Now the inequality becomes: $2x(x+1)^2 - \frac{18(x+1)}{(x-\sqrt{2})(x+\sqrt{2})} \geq 0$

Step 2: Determine the domain of the inequality. The inequality is defined for all real values of $x$ except for the points where the denominator becomes zero. So, the domain is $\mathbb{R} - \{-\sqrt{2}, \sqrt{2}\}$.

Step 3: Find the critical points. The critical points are the values of $x$ that make the expression zero or undefined. In this case, the critical point is $x = -1$, where both terms of the expression become zero.

Step 4: Test intervals and determine the sign of the expression. Let's test three intervals: $(- \infty, -\sqrt{2})$, $(- \sqrt{2}, -1)$, and $(-1, \sqrt{2})$.

1. Test $x = 0$: $2(0)(0+1)^2 - \frac{18(0+1)}{(0-\sqrt{2})(0+\sqrt{2})} = 0 - \frac{18}{-2} = 9 > 0$

2. Test $x = -2$: $2(-2)(-2+1)^2 - \frac{18(-2+1)}{(-2-\sqrt{2})(-2+\sqrt{2})} = -18 - \frac{18}{-4} = -13.5 < 0$

3. Test $x = 1$: $2(1)(1+1)^2 - \frac{18(1+1)}{(1-\sqrt{2})(1+\sqrt{2})} = 8 - \frac{36}{-1} = -44 < 0$

4. Test $x = 3$: $2(3)(3+1)^2 - \frac{18(3+1)}{(3-\sqrt{2})(3+\sqrt{2})} = 80 - \frac{72}{7} > 0$