Question

4- 5х>х+1x2 - х-6<0​

Answer 1

Объяснение:

4-5х>x+1

-5x-x>1-4

-6x>-3

x<3/6

x<1/2

x^2-x-6<0

D=1+24=25

x1=(1+5)/2=3

x2=(1-5)/2=-2

(x-3)(x+3)<0

ну и промежутки сам(а) определи

Answer 2

Let's solve the two inequalities separately:

1. $4 - 5x > x + 1x^2 - x - 6$

First, let's simplify the right-hand side:

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

So the inequality becomes:

$4 - 5x > x^2 - 6$

Now, move all terms to one side to set the inequality to zero:

$x^2 - 5x - 6 - 4 > 0$

Simplify further:

$x^2 - 5x - 10 > 0$

Now, let's find the roots of the quadratic equation:

$x^2 - 5x - 10 = 0$

You can use the quadratic formula to find the roots:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this case, $a = 1$, $b = -5$, and $c = -10$. Plugging these values into the formula:

$x = \frac{5 \pm \sqrt{5^2 - 4(1)(-10)}}{2(1)}$

$x = \frac{5 \pm \sqrt{25 + 40}}{2}$

$x = \frac{5 \pm \sqrt{65}}{2}$

Now, you have two potential values for $x$ from the roots: $x_1 = \frac{5 + \sqrt{65}}{2}$ and (x_2 = \frac{5 - \sqrt{65}}{2}.

Now, we need to test the intervals between these values to determine when the quadratic is greater than zero.

1. Choose a test point in the interval $(-∞, x_2)$, for example, $x = 0$:

   $0^2 - 5(0) - 10 = -10$, which is negative.

2. Choose a test point in the interval $(x_2, x_1)$, for example, $x = 3$:

   $3^2 - 5(3) - 10 = -16$, which is also negative.

3. Choose a test point in the interval $(x_1, +∞)$, for example, $x = 10$:

   $10^2 - 5(10) - 10 = 60$, which is positive.

The inequality is greater than zero in the interval $(x_2, x_1)$, which means the solution to the first inequality is:

$-∞ < x < \frac{5 - \sqrt{65}}{2}$ or $\frac{5 + \sqrt{65}}{2} < x < +∞$

Now, let's solve the second inequality:

2. $x + 1x^2 - x - 6 < 0$

Simplify the right-hand side as we did before:

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

The inequality becomes:

$x^2 - 6 < 0$

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

$x^2 - 6 = 0$

Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$