Question

Решите систему: x^2<=9 x^2+2x+1/x-1

Answer 1

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Answer 2

To solve the given system, we have two inequalities:

1. $x^2 \leq 9$
2. $x^2 + \frac{2x + 1}{x - 1}$

Let's start with the first inequality:

1. $x^2 \leq 9$

To solve this, we'll first find the critical points by setting $x^2$ equal to 9 and then solving for $x$:

$x^2 = 9$

Taking the square root of both sides:

$x = \pm \sqrt{9}$

$x = \pm 3$

Now, we'll create a sign chart to determine the solution regions for $x^2 \leq 9$:

$$\begin{array}{c|cccc} & x < -3 & -3 < x < 3 & x > 3 \\ \hline x^2 - 9 & (-) & (+) & (+) \\ \end{array}$$

From the sign chart, we can see that the inequality $x^2 \leq 9$ is satisfied for $x$ in the range $-3 \leq x \leq 3$. So the solution to the first inequality is $-3 \leq x \leq 3$.

Now let's move on to the second inequality:

2. $x^2 + \frac{2x + 1}{x - 1}$

To solve this inequality, we first need to find the critical points by setting the expression equal to zero and then solving for $x$:

$x^2 + \frac{2x + 1}{x - 1} = 0$

The expression is undefined when $x - 1 = 0$, so $x = 1$ is not in the domain.

Now, we'll find the solutions for the inequality by considering the sign of the expression in different intervals:

$$\begin{array}{c|cccc} & x < 1 & 1 < x < 3 & x > 3 \\ \hline x^2 + \frac{2x + 1}{x - 1} & ? & ? & ? \end{array}$$

To determine the signs, we can pick test points in each interval and evaluate the expression.

For $x = 0$ (in the interval $x < 1$):

$0^2 + \frac{2 \cdot 0 + 1}{0 - 1} = -1$

For $x = 2$ (in the interval $1 < x < 3$):

$2^2 + \frac{2 \cdot 2 + 1}{2 - 1} = 4 + 5 = 9$

For $x = 4$ (in the interval $x > 3$):

$4^2 + \frac{2 \cdot 4 + 1}{4 - 1} = 16 + \frac{9}{3} = 19$

Now, let's complete the sign chart:

$$\begin{array}{c|cccc} & x < 1 & 1 < x < 3 & x > 3 \\ \hline x^2 + \frac{2x + 1}{x - 1} & (-) & (+) & (+) \end{array}$$

From the sign chart, we can see that the expression $x^2 + \frac{2x + 1}{x - 1}$ is positive for $x$ in the intervals $1 < x < 3$ and $x > 3$. It is negative for $x$ in the interval $x < 1$