Question

9 |x| - 2 = 34 5 |x| + 7 = 52 |5x - 3| = 7 |12 - 4x| = 0 |9x - 17| = -21

Answer 1

1.

9|х|-2=34

9|х|=34+2

9|х|=36

|х|=36:9

|х|=4

х1=4

х2= -4

2.

5|х|+7=34

5|х|=34-7

5|х|=27

|х|=27:5

|х|=5,4

х1=5,4

х2= -5,4

3.

|5х-3|=7

1)5х-3=7

5х=7+3

5х=10

х=10:5

х1=2

2)5х-3= -7

5х= -7+3

5х= -4

х= -4:5

х2= -0,8

4.

|12-4х|=0

12-4х=0

4х=12-0

4х=12

х=12:4

х=3

5.

Нет корней так как модуль не может быть отрицательным.

х=∅

Answer 2

To solve absolute value equations, you need to consider both the positive and negative cases for the absolute value expression. Here's how you can solve each of the equations:

1. $9 |x| - 2 = 34$: First, isolate the absolute value term: $9 |x| = 34 + 2$ $9 |x| = 36$

   Now, divide both sides by 9: $|x| = \frac{36}{9}$ $|x| = 4$

   Since the absolute value of a number can be either positive or negative, you have two cases to consider: $x = 4$ or $x = -4$

2. $5 |x| + 7 = 52$: Isolate the absolute value term: $5 |x| = 52 - 7$ $5 |x| = 45$

   Divide both sides by 5: $|x| = \frac{45}{5}$ $|x| = 9$

   Again, consider both cases: $x = 9$ or $x = -9$

3. $|5x - 3| = 7$: Here, you already have the absolute value isolated.

   Now, consider two cases: Case 1: $5x - 3 = 7$ Solve for $x$: $5x = 7 + 3$ $5x = 10$ $x = 2$

   Case 2: $-(5x - 3) = 7$ (negating the absolute value) Solve for $x$: $-5x + 3 = 7$ $-5x = 7 - 3$ $-5x = 4$ $x = -\frac{4}{5}$

   So, $x = 2$ or $x = -\frac{4}{5}$

4. $|12 - 4x| = 0$: The absolute value of any real number is non-negative, and it's only equal to zero when the number inside is zero.

   $12 - 4x = 0$

   Solve for $x$: $4x = 12$ $x = 3$

   There's only one solution: $x = 3$

5. $|9x - 17| = -21$: Absolute values cannot be negative. Since the absolute value is equal to a negative number, there are no real solutions to this equation.

So, the solutions for each equation are:

1. $x = 4$ or $x = -4$
2. $x = 9$ or $x = -9$
3. $x = 2$ or $x = -\frac{4}{5}$
4. $x = 3$
5. No real solutions.