Let $M$ and $N$ , respectively, be the points of the sides $[AB]$ and $[BC]$ of a rectangle $ABCD$ . Let $a,b,c$ be the area of the triangle $AMD,MBN, NCD$ respectively. Express in terms of $a,b$ and $c$ , the area of the triangle $DMN$ .
1. **Define the dimensions and areas:**
   Let the dimensions of the rectangle \(ABCD\) be \(AD = x\) and \(CD = y\). The areas of the triangles are given as:
   - \( \text{Area of } \triangle AMD = a \)
   - \( \text{Area of } \triangle MBN = b \)
   - \( \text{Area of } \triangle NCD = c \)

2. **Express lengths in terms of areas:**
   - The area of \(\triangle AMD\) can be expressed as:
     \[
     a = \frac{1}{2} \times AM \times x \implies AM = \frac{2a}{x}
     \]
   - The area of \(\triangle NCD\) can be expressed as:
     \[
     c = \frac{1}{2} \times CN \times y \implies CN = \frac{2c}{y}
     \]

3. **Express the area of \(\triangle MBN\):**
   - The area of \(\triangle MBN\) can be expressed as:
     \[
     b = \frac{1}{2} \times MB \times BN
     \]
   - Since \(MB = y - \frac{2a}{x}\) and \(BN = x - \frac{2c}{y}\), we have:
     \[
     2b = (y - \frac{2a}{x})(x - \frac{2c}{y})
     \]

4. **Expand and simplify the expression:**
   \[
   2b = yx - y \cdot \frac{2c}{y} - x \cdot \frac{2a}{x} + \frac{4ac}{xy}
   \]
   \[
   2b = xy - 2c - 2a + \frac{4ac}{xy}
   \]

5. **Solve for \(xy\):**
   \[
   2b = xy - 2a - 2c + \frac{4ac}{xy}
   \]
   Multiply both sides by \(xy\):
   \[
   2b \cdot xy = (xy)^2 - 2a \cdot xy - 2c \cdot xy + 4ac
   \]
   \[
   2b \cdot xy = (xy)^2 - 2(a+c) \cdot xy + 4ac
   \]
   Let \(z = xy\):
   \[
   2bz = z^2 - 2(a+c)z + 4ac
   \]
   \[
   z^2 - 2(a+c)z + 4ac - 2bz = 0
   \]
   This is a quadratic equation in \(z\):
   \[
   z^2 - 2(a+c)z + 4ac - 2bz = 0
   \]

6. **Solve the quadratic equation:**
   Using the quadratic formula \(z = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\), where \(A = 1\), \(B = -2(a+c)\), and \(C = 4ac - 2b\):
   \[
   z = \frac{2(a+c) \pm \sqrt{(2(a+c))^2 - 4 \cdot 1 \cdot (4ac - 2b)}}{2 \cdot 1}
   \]
   \[
   z = (a+c) \pm \sqrt{(a+c)^2 - (4ac - 2b)}
   \]
   \[
   z = (a+c) \pm \sqrt{(a+c)^2 - 4ac + 2b}
   \]
   \[
   z = (a+c) \pm \sqrt{(a+c)^2 - 4ac + 2b}
   \]
   \[
   z = (a+c) \pm \sqrt{(a+c)^2 - 4ac + 2b}
   \]

7. **Determine the area of \(\triangle DMN\):**
   The area of \(\triangle DMN\) is:
   \[
   \text{Area of } \triangle DMN = xy - a - b - c
   \]
   Substituting \(xy = z\):
   \[
   \text{Area of } \triangle DMN = z - a - b - c
   \]
   \[
   \text{Area of } \triangle DMN = (a+c) + \sqrt{(a+b+c)^2 - 4ac} - a - b - c
   \]
   \[
   \text{Area of } \triangle DMN = \sqrt{(a+b+c)^2 - 4ac}
   \]

The final answer is \( \boxed{ \sqrt{(a+b+c)^2 - 4ac} } \)