Question

X - 3 < - 2 / x - 2 )))))

Answer 1

x - 3 < - 2 / x - 2

(х²-3х+2+2х)/х<0

(х²-х+2)/х<0

х²-х+2 больше нуля при любом х, т.к. дискриминант равен 1-8=-7 - отрицательный, а первый коэффициент равен 1, положительных.

Значит, при х меньшем нулю выполняется данное неравенство.

__________0_________

    -                      +

Ответ х∈(-∞;0)

Answer 2

Ответ:

Пошаговое объяснение:

Решение на фото

Answer 3

It looks like you have a complex inequality with a lot of parentheses. Let's simplify it step by step.

The given inequality is: X - 3 < -2 / (x - 2)

First, let's get rid of the fraction by multiplying both sides of the inequality by (x - 2), but we need to be careful about the sign of (x - 2):

1. If x - 2 is positive (greater than zero), then we don't need to change the direction of the inequality.
2. If x - 2 is negative (less than zero), we should reverse the direction of the inequality.

Now, we'll consider both cases:

Case 1: (x - 2) is positive (x > 2)

(x - 2) * (x - 3) < -2

Now, let's expand and simplify:

x^2 - 5x + 6 < -2

Subtract -2 from both sides:

x^2 - 5x + 6 - (-2) < 0

x^2 - 5x + 8 < 0

Now, we have a quadratic inequality. We can solve it by finding the critical points, which are the values of x that make the expression equal to zero:

x^2 - 5x + 8 = 0

This quadratic equation doesn't have real solutions because the discriminant (b^2 - 4ac) is negative.

So, for this case (x > 2), there are no real solutions.

Case 2: (x - 2) is negative (x < 2)

In this case, we need to reverse the direction of the inequality:

-(x - 2) * (x - 3) < -2

Now, let's expand and simplify:

-(x^2 - 5x + 6) < -2

Distribute the negative sign:

-x^2 + 5x - 6 < -2

Add 2 to both sides:

-x^2 + 5x - 6 + 2 < 0

-x^2 + 5x - 4 < 0

Now, we have a quadratic inequality. We can solve it by finding the critical points:

x^2 - 5x + 4 = 0

Factor the quadratic:

(x - 4)(x - 1) = 0

Set each factor equal to zero:

1. x - 4 = 0 x = 4

2. x - 1 = 0 x = 1

So, the critical points are x = 1 and x = 4.

Now, we can test the intervals between these critical points:

• When x < 1: -x^2 + 5x - 4 < 0 is satisfied.

• When 1 < x < 4: -x^2 + 5x - 4 < 0 is satisfied.

• When x > 4: -x^2 + 5x - 4 < 0 is not satisfied.

Therefore, the solution to the original inequality is:

x < 1 or 1 < x < 4 (the solution is in the form of intervals).