The least positive value of `x` satisfying the equation `|x+1|-|x|+3|x-1|+2|x-2|= x+2` is

Explanation

To solve the equation \( |x+1| - |x| + 3|x-1| + 2|x-2| = x + 2 \), we will analyze the expression based on the different intervals defined by the points where the absolute values change, namely at \( x = -1, 0, 1, \) and \( 2 \).

### Step 1: Identify the intervals
The critical points are \( x = -1, 0, 1, \) and \( 2 \). We will analyze the equation in the following intervals:
1. \( x < -1 \)
2. \( -1 \leq x < 0 \)
3. \( 0 \leq x < 1 \)
4. \( 1 \leq x < 2 \)
5. \( x \geq 2 \)

### Step 2: Solve for each interval

#### Interval 1: \( x < -1 \)
In this case:
- \( |x+1| = -(x+1) = -x - 1 \)
- \( |x| = -x \)
- \( |x-1| = -(x-1) = -x + 1 \)
- \( |x-2| = -(x-2) = -x + 2 \)

Substituting these into the equation:
\[ -x - 1 - (-x) + 3(-x + 1) + 2(-x + 2) = x + 2 \]
This simplifies to:
\[ -x - 1 + x - 3x + 3 - 2x + 4 = x + 2 \]
Combining like terms gives:
\[ -5x + 6 = x + 2 \]
Rearranging yields:
\[ -6x + 6 = 2 \implies -6x = -4 \implies x = \frac{2}{3} \]
Since \( \frac{2}{3} \) is not in this interval, we discard this solution.

#### Interval 2: \( -1 \leq x < 0 \)
Here:
- \( |x+1| = x + 1 \)
- \( |x| = -x \)
- \( |x-1| = -x + 1 \)
- \( |x-2| = -x + 2 \)

Substituting these into the equation:
\[ (x + 1) - (-x) + 3(-x + 1) + 2(-x + 2) = x + 2 \]
This simplifies to:
\[ x + 1 + x - 3x + 3 - 2x + 4 = x + 2 \]
Combining like terms gives:
\[ -3x + 8 = x + 2 \]
Rearranging yields:
\[ -4x + 8 = 2 \implies -4x = -6 \implies x = \frac{3}{2} \]
Since \( \frac{3}{2} \) is not in this interval, we discard this solution.

#### Interval 3: \( 0 \leq x < 1 \)
In this case:
- \( |x+1| = x + 1 \)
- \( |x| = x \)
- \( |x-1| = -x + 1 \)
- \( |x-2| = -x + 2 \)

Substituting these into the equation:
\[ (x + 1) - x + 3(-x + 1) + 2(-x + 2) = x + 2 \]
This simplifies to:
\[ 1 + 3 - 3x + 4 - 2x = x + 2 \]
Combining like terms gives:
\[ 8 - 5x = x + 2 \]
Rearranging yields:
\[ -6x + 8 = 2 \implies -6x = -6 \implies x = 1 \]
Since \( 1 \) is at the boundary, we will check it later.

#### Interval 4: \( 1 \leq x < 2 \)
Here:
- \( |x+1| = x + 1 \)
- \( |x| = x \)
- \( |x-1| = x - 1 \)
- \( |x-2| = -x + 2 \)

Substituting these into the equation:
\[ (x + 1) - x + 3(x - 1) + 2(-x + 2) = x + 2 \]
This simplifies to:
\[ 1 + 3x - 3 + 4 - 2x = x + 2 \]
Combining like terms gives:
\[ -2 + 2x = x + 2 \]
Rearranging yields:
\[ x - 2 = 2 \implies x = 4 \]
Since \( 4 \) is not in this interval, we discard this solution.

#### Interval 5: \( x \geq 2 \)
In this case:
- \( |x+1| = x + 1 \)
- \( |x| = x \)
- \( |x-1| = x - 1 \)
- \( |x-2| = x - 2 \)

Substituting these into the equation:
\[ (x + 1) - x + 3(x - 1) + 2(x - 2) = x + 2 \]
This simplifies to:
\[ 1 + 3x - 3 + 2x - 4 = x + 2 \]
Combining like terms gives:
\[ 2x - 6 = x + 2 \]
Rearranging yields:
\[ x - 6 = 2 \implies x = 8 \]
Since \( 8 \) is valid in this interval, we keep this solution.

### Final Check
The least positive value satisfying the equation is \( x = 1 \).

### Conclusion
The least positive value of \( x \) satisfying the equation is:
\[ \boxed{1} \]