Question

Tìm giá trị nhỏ nhất của biểu thức: \(B=x+7-\sqrt{x-5}\)

Answer 1

ĐK:\(\sqrt{x-5}\ge0\Leftrightarrow x-5\ge0\Leftrightarrow x\ge5\)
Ta có:
\(B=x+7-\sqrt{x-5}\)
\(=x-5-\sqrt{x-5}+12\)
\(=x-5-2.\sqrt{x-5}.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{47}{4}\)
\(=\left(\sqrt{x-5}-\dfrac{1}{4}\right)^2+\dfrac{47}{2}\ge\dfrac{47}{2}\)
suy ra GTNN của B=\(\dfrac{47}{2}\Leftrightarrow\left(\sqrt{x-5}-\dfrac{1}{2}\right)^2=0\Leftrightarrow\sqrt{x-5}=\dfrac{1}{2}\)
\(\Leftrightarrow x-5=\dfrac{1}{4}\Leftrightarrow x=\dfrac{21}{4}\left(tm\right)\)

Answer 2

\(x\ge5\)

\(B=x-5-\sqrt{x-5}+\dfrac{1}{4}+\dfrac{47}{4}=\left(\sqrt{x-5}-\dfrac{1}{2}\right)^2+\dfrac{47}{4}\ge\dfrac{47}{4}\)

\(\Rightarrow B_{min}=\dfrac{47}{4}\) khi \(\sqrt{x-5}-\dfrac{1}{2}=0\Rightarrow x=\dfrac{21}{4}\)