Question

||х|-5|=1||х|-2|=5||х|-3|=3​

Answer 1

Ответ:

Пошаговое объяснение:

llxl - 5l = 1

lxl - 5 = 1 lxl - 5 = - 1

lxl = 6 lxl = 4

x = 6 x = - 6 x = 4 x = - 4

Ответ : х₁ = - 6 ; х₂ = - 4 ; х₃ = 4 ; х₄ = 6

llxl - 2l = 5

lxl - 2 = 5 lxl - 2 = - 5

lxl = 7 lxl = - 3

x = 7 x = - 7 x∈∅

Ответ : х₁ = - 7 ; х = 7

llxl - 3l = 3

lxl - 3 = 3 lxl - 3 = - 3

lxl = 6 lxl = 0

x = 6 x = - 6 x = 0

Ответ : х₁ = - 6 ; х₂ = 0 ; х₃ = 6

Answer 2

To solve this system of absolute value equations:

1. |x - 5| = 1
2. |x - 2| = 5
3. |x - 3| = 3

You can break it down into individual equations and solve for x in each case:

Equation 1: |x - 5| = 1

There are two possible cases for this equation:

Case 1: (x - 5) = 1 x - 5 = 1 x = 1 + 5 x = 6

Case 2: -(x - 5) = 1 -x + 5 = 1 -x = 1 - 5 -x = -4 x = 4 (Note that we changed the sign because it's the negative of the absolute value of x - 5)

So, for Equation 1, you have two solutions: x = 6 and x = 4.

Equation 2: |x - 2| = 5

Again, there are two possible cases for this equation:

Case 1: (x - 2) = 5 x - 2 = 5 x = 5 + 2 x = 7

Case 2: -(x - 2) = 5 -x + 2 = 5 -x = 5 - 2 -x = 3 x = -3

So, for Equation 2, you have two solutions: x = 7 and x = -3.

Equation 3: |x - 3| = 3

Once more, there are two possible cases for this equation:

Case 1: (x - 3) = 3 x - 3 = 3 x = 3 + 3 x = 6

Case 2: -(x - 3) = 3 -x + 3 = 3 -x = 3 - 3 -x = 0 x = 0

So, for Equation 3, you have two solutions: x = 6 and x = 0.

Now, you have found all the solutions for each equation. The solutions for the original system of equations are:

x = 6, 4, 7, -3, 0