Question

1) |x+4|=|x-4|2) |x^2 -5x+6|> x^2 -5x+6

Answer 1

$|x+4|=|x-4| \\\ \left \{ {{x+4=x-4} \atop {x+4=4-x}} \right. \\\ \left \{ {{4 \neq -4} \atop {2x=0}} \right. \\\ x=0$
Ответ: 0

$|x^2 -5x+6|> x^2 -5x+6 \\\ \left \{ {{x^2 -5x+6> x^2 -5x+6} \atop {x^2 -5x+6<- x^2 +5x-6}} \right. \\\ \left \{ {{0>0} \atop {2x^2 -10x+12<0}} \right. \\\ x^2 -5x+6<0 \\\ D=25-24=1 \\\ x_1= \frac{5+1}{2}=3 \\\ x_2= \frac{5-1}{2}=2 \\\ x\in (2;3)$
Ответ: 2<x<3

Answer 2

1) |x + 4| = |x - 4|

To solve this equation, we need to consider two cases: when the expression inside the absolute value is positive and when it is negative.

Case 1: (x + 4) is positive, (x - 4) is positive In this case, the equation becomes: x + 4 = x - 4

We can see that the variable x cancels out, leaving us with: 4 = -4

Since the equation is inconsistent, there are no solutions in this case.

Case 2: (x + 4) is negative, (x - 4) is negative In this case, the equation becomes: -(x + 4) = -(x - 4)

We can distribute the negative sign: -x - 4 = -x + 4

The variable x cancels out again, and we have: -4 = 4

Once again, the equation is inconsistent, and there are no solutions in this case.

Therefore, the given equation |x + 4| = |x - 4| has no solutions.

2) |x^2 - 5x + 6| > x^2 - 5x + 6

To solve this inequality, we need to consider two cases: when the expression inside the absolute value is positive and when it is negative.

Case 1: (x^2 - 5x + 6) is positive In this case, the inequality becomes: x^2 - 5x + 6 > x^2 - 5x + 6

The variable x cancels out, and we have: 0 > 0

Since the inequality is inconsistent, there are no solutions in this case.

Case 2: (x^2 - 5x + 6) is negative In this case, the inequality becomes: -(x^2 - 5x + 6) > x^2 - 5x + 6

We can distribute the negative sign: -x^2 + 5x - 6 > x^2 - 5x + 6

Combine like terms: -2x^2 + 10x - 12 > 0

To solve this quadratic inequality, we can find the roots of the corresponding quadratic equation: -2x^2 + 10x - 12 = 0

Factoring, we get: -2(x - 2)(x - 3) = 0

Setting each factor to zero, we find the critical points: x - 2 = 0 -> x = 2 x - 3 = 0 -> x = 3

We can now create a sign chart to determine the intervals that satisfy the inequality:

Interval  | -2x^2 + 10x - 12 > 0
-------------------------------
(-∞, 2)   |    +   
(2, 3)    |    -
(3, ∞)    |    +

From the sign chart, we can see that the inequality is satisfied when x is in the intervals (-∞, 2) and (3, ∞).

Therefore, the solution to the given inequality |x^2 - 5x + 6| > x^2 - 5x + 6 is x ∈ (-∞, 2) U (3, ∞).