Question

|x^2-2x| < 3 Полное решение

Answer 1

х²-2х<3; х²-2х≥0

-(х²-2х)<3; х²-2х<0

х принадлежит (-1;3); х принадл. (-бесконечности;0] в объединение [2;+бесконеч)

х принадл R;хпринадл (0;2)

х принадл (-1;3)

Answer 2

To solve the inequality |x^2 - 2x| < 3, we'll break it down into two cases based on the expression inside the absolute value.

Case 1: (x^2 - 2x) < 3 To solve this case, we need to find the values of x that satisfy the inequality (x^2 - 2x) < 3.

Step 1: Subtract 3 from both sides of the inequality: x^2 - 2x - 3 < 0

Step 2: Factor the quadratic expression: (x - 3)(x + 1) < 0

Step 3: Find the critical points where (x - 3)(x + 1) = 0: x - 3 = 0 --> x = 3 x + 1 = 0 --> x = -1

Step 4: Create a sign chart to analyze the inequality: Intervals: (-∞, -1) (-1, 3) (3, ∞) Test Value: x = 0 x = 1 x = 4 Sign of (x - 3)(x + 1): - + +

Step 5: Determine the solution in each interval: (x - 3)(x + 1) < 0 in the interval (-1, 3).

So, the solution for Case 1 is: -1 < x < 3.

Case 2: -(x^2 - 2x) < 3 For this case, we need to solve the inequality -(x^2 - 2x) < 3.

Step 1: Distribute the negative sign: -x^2 + 2x < 3

Step 2: Move all terms to the left-hand side of the inequality: -x^2 + 2x - 3 < 0

Step 3: Factor the quadratic expression: -(x - 3)(x + 1) < 0

Step 4: Create a sign chart to analyze the inequality: Intervals: (-∞, -1) (-1, 3) (3, ∞) Test Value: x = 0 x = 1 x = 4 Sign of -(x - 3)(x + 1): + - +

Step 5: Determine the solution in each interval: -(x - 3)(x + 1) < 0 in the intervals (-∞, -1) and (3, ∞).

So, the solution for Case 2 is: x < -1 or x > 3.

Now, we need to combine the solutions from both cases:

-1 < x < 3 (from Case 1) and x < -1 or x > 3 (from Case 2).

The final solution to the inequality |x^2 - 2x| < 3 is: x < -1 or -1 < x < 3 or x > 3.