If `0ltxltpi/4andcosx+sinx=5/4`, then the value of `16(cosx-sinx)^2` is ___________ .

Explanation

To solve the problem, we need to find the value of \( 16(\cos x - \sin x)^2 \) given that \( \cos x + \sin x = \frac{5}{4} \) and \( 0 < x < \frac{\pi}{4} \).

### Step-by-Step Solution:

1. **Let \( y = \cos x - \sin x \)**:
We will express \( y \) in terms of \( \cos x \) and \( \sin x \).

2. **Square \( y \)**:
\[ y^2 = (\cos x - \sin x)^2 = \cos^2 x - 2\cos x \sin x + \sin^2 x \]
Using the Pythagorean identity \( \cos^2 x + \sin^2 x = 1 \):
\[ y^2 = 1 - 2\cos x \sin x \tag{1} \]

3. **Square the given equation \( \cos x + \sin x = \frac{5}{4} \)**:
\[ (\cos x + \sin x)^2 = \left(\frac{5}{4}\right)^2 \]
Expanding the left side:
\[ \cos^2 x + 2\cos x \sin x + \sin^2 x = \frac{25}{16} \]
Again using \( \cos^2 x + \sin^2 x = 1 \):
\[ 1 + 2\cos x \sin x = \frac{25}{16} \]

4. **Solve for \( 2\cos x \sin x \)**:
\[ 2\cos x \sin x = \frac{25}{16} - 1 = \frac{25}{16} - \frac{16}{16} = \frac{9}{16} \tag{2} \]

5. **Substitute equation (2) into equation (1)**:
\[ y^2 = 1 - 2\cos x \sin x = 1 - \frac{9}{16} \]
Converting 1 to a fraction:
\[ 1 = \frac{16}{16} \]
Thus:
\[ y^2 = \frac{16}{16} - \frac{9}{16} = \frac{7}{16} \]

6. **Find \( 16(\cos x - \sin x)^2 \)**:
Since \( y^2 = \frac{7}{16} \):
\[ 16y^2 = 16 \cdot \frac{7}{16} = 7 \]

### Final Answer:
The value of \( 16(\cos x - \sin x)^2 \) is \( \boxed{7} \).