Question

(x²-x-6)/(x²-2x+1)≤0​

Answer 1

Ответ смотри во вложении

Answer 2

Ответ:

х=6

($6^{2}$-6-6)/($6^{2}$-2*6+1) = 0,96 (Это если десятичными)

Answer 3

To solve the inequality $\frac{x^2 - x - 6}{x^2 - 2x + 1} \leq 0$, we can follow these steps:

1. First, find the critical points by setting the numerator and denominator equal to zero and solving for $x$:

   $$\begin{align*} \text{Numerator:} \quad & x^2 - x - 6 = 0 \\ & (x - 3)(x + 2) = 0 \\ & x_1 = 3, \quad x_2 = -2 \\ \\ \text{Denominator:} \quad & x^2 - 2x + 1 = 0 \\ & (x - 1)^2 = 0 \\ & x_3 = 1 \end{align*}$$

2. These critical points divide the real number line into four intervals: $(- \infty, -2)$, $(-2, 1)$, $(1, 3)$, and $(3, \infty)$.

3. Now, we need to test a point from each interval in the original inequality to determine the sign of the expression. We can use the test points $x = -3, 0, 2, 4$:

   • For $x = -3$: $\frac{(-3)^2 - (-3) - 6}{(-3)^2 - 2(-3) + 1} = \frac{6}{16} = \frac{3}{8} > 0$.
   • For $x = 0$: $\frac{(0)^2 - 0 - 6}{(0)^2 - 2(0) + 1} = \frac{-6}{1} = -6 < 0$.
   • For $x = 2$: $\frac{(2)^2 - 2 - 6}{(2)^2 - 2(2) + 1} = \frac{-2}{1} = -2 < 0$.
   • For $x = 4$: $\frac{(4)^2 - 4 - 6}{(4)^2 - 2(4) + 1} = \frac{6}{9} = \frac{2}{3} > 0$.

4. Analyzing the signs of the expression for these test points, we can determine the intervals where the inequality is satisfied:

   • $(- \infty, -2)$: The expression is positive.
   • $(-2, 1)$: The expression is negative.
   • $(1, 3)$: The expression is negative.
   • $(3, \infty)$: The expression is positive.

5. Now, let's determine where the inequality $\frac{x^2 - x - 6}{x^2 - 2x + 1} \leq 0$ is true. It's true in the intervals where the expression is less than or equal to zero, which are $(-2, 1)$ and $(3, \infty)$.

So, the solution to the inequality is:

$x \in (-2, 1] \cup (3, \infty)$

This represents the values of $x$ for which the inequality is satisfied.