Let A and B be acute angles measured counterclockwise from the...

From the given coordinates, the tangent of the angles can be found using $\tan \theta = \dfrac{y}{x}$.

For angle $A$, the point is $P(2x+1, 1)$, so:
$\tan A = \dfrac{1}{2x+1}$

For angle $B$, the point is $Q(x+1, x)$, so:
$\tan B = \dfrac{x}{x+1}$

Using the tangent addition formula:
$\tan(A+B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}$

Substitute the values of $\tan A$ and $\tan B$:
$\tan(A+B) = \dfrac{\dfrac{1}{2x+1} + \dfrac{x}{x+1}}{1 - \left(\dfrac{1}{2x+1}\right)\left(\dfrac{x}{x+1}\right)}$

$\Rightarrow \tan(A+B) = \dfrac{\dfrac{(x+1) + x(2x+1)}{(2x+1)(x+1)}}{\dfrac{(2x+1)(x+1) - x}{(2x+1)(x+1)}}$

$\Rightarrow \tan(A+B) = \dfrac{x + 1 + 2x^2 + x}{2x^2 + 3x + 1 - x}$

$\Rightarrow \tan(A+B) = \dfrac{2x^2 + 2x + 1}{2x^2 + 2x + 1} = 1$

Since $x > 0$, we have $0 < \tan A < 1$ and $0 < \tan B < 1$, which means $0 < A < \dfrac{\pi}{4}$ and $0 < B < \dfrac{\pi}{4}$.

Therefore, $0 < A+B < \dfrac{\pi}{2}$.

Since $\tan(A+B) = 1$ and $A+B$ is in the first quadrant, we get $A+B = \dfrac{\pi}{4}$.

Answer: $\dfrac{\pi}{4}$