\(A=1+\sqrt{x-2}\)
Do \(\sqrt{x-2}\ge0\forall x>2\) nên \(A\ge1\forall x>2\)
Vậy \(minA=1\Leftrightarrow x=2\)
__________
\(B=5-\sqrt{2x-1}\)
Do \(\sqrt{2x-1}\ge0\forall x\ge\frac{1}{2}\)nên \(B\le5\forall x\ge\frac{1}{2}\)
Vậy \(maxB=5\Leftrightarrow x=\frac{1}{2}\)