Question

[minh]

tìm GTNN của biểu thức \(x-\sqrt{x-3}+4\)

Answer 1

[Nguyễn Đức Trí]

\(A=x-\sqrt[]{x-3}+4\)

\(\Rightarrow A=x-3-\sqrt[]{x-3}+\dfrac{1}{4}-\dfrac{1}{4}-3+4\)

\(\Rightarrow A=\left(\sqrt[]{x-3}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

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 \(GTNN\left(A\right)=\dfrac{3}{4}\left(tạix=\dfrac{13}{4}\right)\)