ĐKXĐ : x≥0;x≠4
Ta có \(\frac{1}{P}=1:\frac{\sqrt{x}+1}{\sqrt{x}-2}=\frac{\sqrt{x}-2}{\sqrt{x}+1}=1-\frac{3}{\sqrt{x}+1}\)
Vơi x≥0;x≠4
=>\(\sqrt{x}\) ≥0
=>\(\sqrt{x}+1\)≥1
=>\(\frac{1}{\sqrt{x}+1}\)≤1
<=>\(\frac{-3}{\sqrt{x}+1}\)≥-3
=>1-\(\frac{3}{\sqrt{x}+1}\)≥-2
=>P≥-2
Dấu "=" xảy ra <=> x=0
Vậy Min P=-2 khi x=0
ĐKXĐ : \(x\ge0;x\ne4\)
Ta có :
\(\frac{1}{P}=\frac{\sqrt{x}-2}{\sqrt{x}+1}=\frac{\sqrt{x}+1-3}{\sqrt{x}+1}=1-\frac{3}{\sqrt{x}+1}\)
Nhận xét :
\(\sqrt{x}\ge0\)
\(\Leftrightarrow\) \(\sqrt{x}+1\ge1\)
\(\Leftrightarrow\) \(\frac{3}{\sqrt{x}+1}\le\frac{3}{1}=3\)
\(\Leftrightarrow\) \(-\frac{3}{\sqrt{x}+1}\ge-3\)
\(\Leftrightarrow\) \(1-\frac{3}{\sqrt{x}+1}\ge1-3=-2\)
Dấu " = " xảy ra khi \(\sqrt{x}+1=1\)\(\Leftrightarrow\) \(x=0\)
Vậy min \(\frac{1}{P}=-2\) khi x = 0
