\(A=x-\sqrt[]{x-3}+4\)
\(\Rightarrow A=x-3-\sqrt[]{x-3}+\dfrac{1}{4}-\dfrac{1}{4}-3+4\)
\(\Rightarrow A=\left(\sqrt[]{x-3}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
mà \(\left(\sqrt[]{x-3}-\dfrac{1}{2}\right)^2\ge0,\forall x\ge3\)
\(\Rightarrow A=\left(\sqrt[]{x-3}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Dấu "=" xảy ra khi và chỉ khi
\(\sqrt[]{x-3}-\dfrac{1}{2}=0\)
\(\Leftrightarrow\sqrt[]{x-3}=\dfrac{1}{2}\)
\(\Leftrightarrow x-3=\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{1}{4}+3\)
\(\Leftrightarrow x=\dfrac{13}{4}\)
Vậy \(GTNN\left(A\right)=\dfrac{3}{4}\left(tạix=\dfrac{13}{4}\right)\)