\(\dfrac{4x-3}{x-1}< 5\)

\(\Leftrightarrow4x-3< 5\left(x-1\right)\)

\(\Leftrightarrow4x-3< 5x-5\)

\(\Leftrightarrow4x-3-5x+5< 0\)

\(\Leftrightarrow-x+2< 0\)

\(\Leftrightarrow-x< -2\)

\(\Leftrightarrow x>2\)

Vậy x > 2

\(=\sqrt{\left(3\sqrt{2}-1\right)^2}-\sqrt{\left(2\sqrt{2}-1\right)^2}\)

\(=3\sqrt{2}-1-2\sqrt{2}+1=\sqrt{2}\)