Find the value of x.

Solution:

Let the triangle be ABC, where the right angle is at vertex C. Let the point on AC be D such that AD = 2 and DC = 3. Let the vertex B be opposite to the hypotenuse. So we have a right-angled triangle ABC, with a point D on AC. The lengths given are AD = 2, DC = 3. The length of the hypotenuse from A to B is given as x-?. The length of the segment DB is given as 4. The length of BC is unknown. Let BC = h.

In triangle DBC, by Pythagorean theorem:

\( DB^2 = DC^2 + BC^2 \)

\( 4^2 = 3^2 + h^2 \)

\( 16 = 9 + h^2 \)

\( h^2 = 16 - 9 \)

\( h^2 = 7 \)

\( h = \sqrt{7} \)

Now, consider the larger right-angled triangle ABC. The length of AC is AD + DC = 2 + 3 = 5. The length of BC is h = \( \sqrt{7} \).

By Pythagorean theorem in triangle ABC:

\( AB^2 = AC^2 + BC^2 \)

The length of AB is given as x-?. Let's assume the value next to x is 2 as it is not clearly visible in the image. So, AB = x-2.

\[ (x-2)^2 = 5^2 + (\sqrt{7})^2 \]

\[ (x-2)^2 = 25 + 7 \]

\[ (x-2)^2 = 32 \]

Taking the square root of both sides:

\[ x-2 = \sqrt{32} \]

\[ x-2 = 4\sqrt{2} \]

\[ x = 2 + 4\sqrt{2} \]

Note: The value next to 'x-' in the image is not entirely clear. Assuming it to be '2'. If the value is different, the final answer for 'x' will change accordingly.

Ответ: x = 2 + 4√2