Question

1)x^2-6x+5>02)(x-8)(x+4)(x-3)<03)9x^2-6x+1>04)2x^2+3x-5<0​

Answer 1

Ответ:

Пошаговое объяснение:

х²-6х+5>0

Корни:

х₁=5 ;   х₂=1.

(х-5)(х-1)>0

++++++++(1)----------(5)++++++

х∈(-∞;1)∪(5;+∞).

2)

(х-8)(х+4)(х-3)<0

Корни:

х₁=8; х₂=-4; х₃=3

-----(-)-----(-4)++(+)+++(3)-----(-)-------(8)+++(+)++

х∈(-∞;-4)∪(3;8).

3)9х²-6х+1>0

Корни: х₁=4/9 ; х₂=2/9.

++++++(2/9)-----------(4/9)++++++

х∈(-∞;2/9)∪(4/9; +∞).

4)

2х²+3х-5<0

Корни:

х₁=1 ; х₂=-5/2= - 2,5

2(х-1)(х+2,5)<0

(х-1)(2х+5)<0

+++++++(-2,5)-------------(1)+++++++

х∈(-2,5; 1 ).

Answer 2

Let's solve each of these inequalities one by one:

1. $x^2 - 6x + 5 > 0$:

To solve this quadratic inequality, we first find the roots of the corresponding equation:

$x^2 - 6x + 5 = 0$

Factoring:

$(x - 5)(x - 1) = 0$

So, the roots are $x = 5$ and $x = 1$.

Now, we can use these roots to determine the intervals where the inequality is satisfied. We create a sign chart:

$$\begin{array}{|c|c|c|c|c|c|} \hline \text{Interval} & (-\infty, 1) & (1, 5) & (5, \infty) \\ \hline x^2 - 6x + 5 & + & - & + \\ \hline \end{array}$$

In the interval $(-\infty, 1)$, $x^2 - 6x + 5$ is positive. In the interval $(1, 5)$, $x^2 - 6x + 5$ is negative. In the interval $(5, \infty)$, $x^2 - 6x + 5$ is positive.

So, the solution to the inequality $x^2 - 6x + 5 > 0$ is $x \in (-\infty, 1) \cup (5, \infty)$.

2. $(x - 8)(x + 4)(x - 3) < 0$:

To solve this inequality, we first find the roots of the corresponding equation:

$(x - 8)(x + 4)(x - 3) = 0$

The roots are $x = 8$, $x = -4$, and $x = 3$.

Now, we create a sign chart:

$$\begin{array}{|c|c|c|c|c|c|c|} \hline \text{Interval} & (-\infty, -4) & (-4, 3) & (3, 8) & (8, \infty) \\ \hline (x - 8)(x + 4)(x - 3) & - & + & - & + \\ \hline \end{array}$$

In the interval $(-\infty, -4)$, $(x - 8)(x + 4)(x - 3)$ is negative. In the interval $(-4, 3)$, $(x - 8)(x + 4)(x - 3)$ is positive. In the interval $(3, 8)$, $(x - 8)(x + 4)(x - 3)$ is negative. In the interval $(8, \infty)$, $(x - 8)(x + 4)(x - 3)$ is positive.

So, the solution to the inequality $(x - 8)(x + 4)(x - 3) < 0$ is $x \in (-\infty, -4) \cup (3, 8)$.

3. $9x^2 - 6x + 1 > 0$:

To solve this quadratic inequality, we first find the roots of the corresponding equation:

$9x^2 - 6x + 1 = 0$

Factoring:

$(3x - 1)(3x - 1) = 0$

So, the root is $x = \frac{1}{3}$.

Now, we create a sign chart:

$$\begin{array}{|c|c|c|c|} \hline \text{Interval} & (-\infty, \frac{1}{3}) & (\frac{1}{3}, \infty) \\ \hline 9x^2 - 6x + 1 & + & + \\ \hline \end{array}$$