[tex] \frac{( \sqrt{x) {}^{2} } - 2 - x {}^{2} }{ x + 9} \leqslant 0[/tex]

Предыдущий
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вопрос

To solve this inequality, we first simplify the expression inside the brackets:

[tex] \frac{\sqrt{x}^2 - 2 - x^2}{x + 9} [/tex]

Simplifying further, we get:

[tex] \frac{x - 2 - x^2}{x + 9} [/tex]

[tex] \frac{-x^2 + x - 2}{x + 9} [/tex]

Now, we need to find the values of x for which this expression is less than or equal to 0. We can do this by finding the critical points, which are the points where the expression is equal to 0.

Setting the expression equal to 0:

[tex] -x^2 + x - 2 = 0 [/tex]

Solving this quadratic equation, we get:

[tex] x = \frac{-1 \pm \sqrt{1 + 8}}{-2} [/tex]

[tex] x = \frac{-1 \pm \sqrt{9}}{-2} [/tex]

[tex] x = \frac{-1 \pm 3}{-2} [/tex]

Thus, the critical points are x = -2 and x = 1.

Now, we need to test the intervals created by these critical points to determine when the expression is less than or equal to 0.

When x < -2:

Choosing a test point x = -3, we get:

[tex] \frac{-(-3)^2 - 3 - 2}{-3 + 9} = \frac{-9 - 3 - 2}{6} = \frac{-14}{6} < 0 [/tex]

When -2 < x < 1:

Choosing a test point x = 0, we get:

[tex] \frac{0 - 2 - 0}{0 + 9} = \frac{-2}{9} < 0 [/tex]

When x > 1:

Choosing a test point x = 2, we get:

[tex] \frac{-2^2 + 2 - 2}{2 + 9} = \frac{-4 + 2 - 2}{11} = \frac{-4}{11} < 0 [/tex]

Therefore, the solution to the inequality is:

[tex] x \in (-2, 1] [/tex]