Question

|x-3|(x^2-(a+2)x+2a)>0

Answer 1

Ответ:

3+2x^2-x-2=0

(x^3+2x^2)-(x+2)=0

x^2*(x+2)-(x+2)=0

(x+2)*(x^2-1)=0

1)x+2=0

x=-2

2)x^2-1=0

x^2=1

x1=1

x2=-1

Ответ ----- (х1=1,х2=-1,х3=-2)

Answer 2

To solve the inequality |x - 3|(x^2 - (a + 2)x + 2a) > 0, we can break it down into different intervals based on the sign of the expression |x - 3| and the expression (x^2 - (a + 2)x + 2a).

First, let's analyze the expression |x - 3|. This expression is equal to x - 3 if x ≥ 3, and it's equal to -(x - 3) if x < 3.

Next, let's analyze the expression (x^2 - (a + 2)x + 2a). We can factor this expression as follows:

x^2 - (a + 2)x + 2a = (x - a)(x - 2)

Now, we'll consider different cases:

Case 1: x < 3 In this case, |x - 3| = -(x - 3), and we have: -(x - 3)(x - a)(x - 2) > 0

This inequality holds true when either all three factors are negative or exactly one factor is negative. Solving for the intervals:

x - 3 < 0 (since x < 3) x - a > 0 (since x < a, as x < 3) x - 2 > 0 (since x < 2)

These inequalities are satisfied when: x < 3, x < a, x > 2.

Case 2: x ≥ 3 In this case, |x - 3| = x - 3, and we have: (x - 3)(x - a)(x - 2) > 0

This inequality holds true when either all three factors are positive or exactly one factor is positive. Solving for the intervals:

x - 3 > 0 (since x ≥ 3) x - a < 0 (since x ≥ a, as x ≥ 3) x - 2 < 0 (since x ≥ 2)

These inequalities are satisfied when: x > 3, x < a, x < 2.

Now, combining the conditions from both cases: For the expression to be greater than 0, the conditions must hold in both cases:

x < 3 and x < a and x > 2

However, these conditions cannot be satisfied simultaneously, as requiring x to be both less than 2 and greater than 3 is a contradiction. This means that there are no values of x that satisfy the given inequality for any value of a.

In conclusion, the given inequality |x - 3|(x^2 - (a + 2)x + 2a) > 0 has no solutions for any real values of x and a.