Question

tìm gtnn của \(\dfrac{x-4}{\sqrt{x}+1}\)

Answer 1

\(ĐKXĐ:x\ge0\)

- Đặt \(A=\dfrac{x-4}{\sqrt{x}+1}\)

\(A=\dfrac{x-4}{\sqrt{x}+1}=\dfrac{\left(x-1\right)-3}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x+1}\right)\left(\sqrt{x}-1\right)-3}{\sqrt{x}+1}=\sqrt{x}-1-\dfrac{3}{\sqrt{x}+1}\)

- Vì \(\sqrt{x}\ge0\Leftrightarrow\dfrac{3}{\sqrt{x}+1}\le\dfrac{3}{1}=3\Leftrightarrow-\dfrac{3}{\sqrt{x}+1}\ge-3\)

- Vậy \(A\ge0-1-3=-4\)

- Kết luận: \(MinA=-4\), đạt tại \(x=0\)