Dat
\(A=\frac{x+8}{\sqrt{x}+1}\).\(DK:x\ge0\)
\(\Leftrightarrow-x+A\sqrt{x}+A-8=0\)
\(\Leftrightarrow x-A\sqrt{x}-A+8=0\)
Dat \(t=\sqrt{x}\left(t\ge0\right)\)
\(\Rightarrow t^2-At-A+8=0\)
Ta co:
\(\Delta\ge0\)
\(\Leftrightarrow A^2+4A-32\ge0\)
\(\Leftrightarrow A^2-4A+8A-32\ge0\)
\(\Leftrightarrow\left(A-4\right)\left(A+8\right)\ge0\)
Vì tìm GTNN nên xét 1TH thôi nhé
\(\Leftrightarrow\left\{{}\begin{matrix}A\ge4\left(1\right)\\A\ge-8\left(2\right)\end{matrix}\right.\)
(1)Dau '=' xay ra khi \(t=-\frac{A}{2}=-2\left(l\right)\).Vi \(t\ge0\)
(2)Dau '=' xay ra khi \(t=-\frac{A}{2}=4\Rightarrow x=16\left(n\right)\)
Vay \(A_{min}=-8\)khi \(x=16\)
ĐK: x≥0x≥0
A=x+8√x+1=(x−1)+9√x+1=√x−1+9√x+1A=x+8x+1=(x−1)+9x+1=x−1+9x+1
=(√x+1+9√x+1)−2≥2.√(√x+1).9√x+1−2=4=(x+1+9x+1)−2≥2.(x+1).9x+1−2=4 (BĐT AM−GMAM−GM )
Vậy GTNN của A là 4 khi x=4
ĐK: x≥0
A= \(\frac{x+8}{\sqrt{x}+1}=\frac{\left(x-1\right)+9}{\sqrt{x}+1}=\sqrt{x}-1+\frac{9}{\sqrt{x}+1}=\left(\sqrt{x}-1+\frac{9}{\sqrt{x}+1}\right)-2\ge2\sqrt{\frac{\left(\sqrt{x}+1\right).9}{\sqrt{x}+1}}-2=4\left(BĐTAM-GM\right)\)