Question

5|4+2(х-3)|=1 модуль решения

Answer 1

Смотри решение в фото

Answer 2

To solve the equation |5|4+2(х-3)|=1|, we need to consider two cases since we have a modulus on the left-hand side. Remember that the modulus of a number is always non-negative (it gives the distance from 0 on the number line). Therefore, the expression inside the modulus must be either 1 or -1 for the whole equation to be satisfied.

Case 1: |4+2(х-3)| = 1 In this case, we can eliminate the modulus since the expression inside it must be either 1 or -1.

1. 4 + 2(х - 3) = 1
2. Solve for x: 2(х - 3) = 1 - 4 2(х - 3) = -3 х - 3 = -3/2 х = -3/2 + 3 х = -3/2 + 6/2 х = (6 - 3) / 2 х = 3/2

Case 2: |4+2(х-3)| = -1 Now, since the modulus result cannot be negative, there are no solutions in this case.

Therefore, the solution to the equation |5|4+2(х-3)|=1| is x = 3/2.