Question

Решить x^2-3x+2/x^2+3x+2≥1

Answer 1

$\frac{ {x}^{2} - 3x + 2 - {x}^{2} - 3x - 2 }{(x + 1)(x + 2)} \geqslant0 \\ - \frac{6x}{(x + 1)(x + 2)} \geqslant 0$
х=-2
х=-1
х=0
__+_-2__-_-1___+_0__-___
(-&;-2)(-1;0]

Answer 2

To solve the inequality $\frac{x^2-3x+2}{x^2+3x+2} \geq 1$, we need to follow these steps:

Step 1: Find the common denominator. Step 2: Simplify the inequality. Step 3: Solve for $x$.

Let's begin:

Step 1: Find the common denominator: The denominators are already in factored form: $x^2 + 3x + 2 = (x+1)(x+2)$

Step 2: Simplify the inequality: $\frac{x^2-3x+2}{x^2+3x+2} \geq 1$

Since the numerator and denominator have the same factor, we can cancel them out:

$\frac{x^2-3x+2}{(x+1)(x+2)} \geq 1$

Step 3: Solve for $x$:

To solve the inequality, we need to find the critical points where the expression is equal to 1 and where it changes sign.

$\frac{x^2-3x+2}{(x+1)(x+2)} = 1$

Multiply both sides by $(x+1)(x+2)$ to eliminate the fraction:

$x^2 - 3x + 2 = (x+1)(x+2)$

Expand and simplify:

$x^2 - 3x + 2 = x^2 + 3x + 2$

Subtract $x^2$ from both sides:

$-3x + 2 = 3x + 2$

Subtract $3x$ from both sides:

$2 = 6x$

Divide by 6:

$x = \frac{1}{3}$

Now, we have a critical point at $x = \frac{1}{3}$. This means the expression $\frac{x^2-3x+2}{(x+1)(x+2)}$ changes sign around this point.

Let's check the sign of the expression in different intervals:

1. $x < -2$: Choose $x = -3$, which is less than -2: $\frac{(-3)^2 - 3(-3) + 2}{((-3) + 1)((-3) + 2)} = \frac{18}{12} > 1$

2. $-2 < x < -1$: Choose $x = -\frac{3}{2}$, which is between -2 and -1: $\frac{\left(-\frac{3}{2}\right)^2 - 3\left(-\frac{3}{2}\right) + 2}{\left(-\frac{3}{2} + 1\right)\left(-\frac{3}{2} + 2\right)} = \frac{\frac{1}{4} + \frac{9}{2} + 2}{\frac{1}{2} \cdot \frac{1}{2}} = \frac{2}{1} = 2 > 1$

3. $-1 < x < \frac{1}{3}$: Choose $x = 0$, which is between -1 and $\frac{1}{3}$: $\frac{0^2 - 3(0) + 2}{(0 + 1)(0 + 2)} = \frac{2}{2} = 1$

4. $x>\frac{1}{3}$ 0 0