Tìm giá trị nhỏ nhất

K

kisihoangtoc

[TEX]\frac{x+8}{\sqrt[]{x}+1}=\frac{x-1}{\sqrt[]{x}+1}+\frac{9}{\sqrt[]{x}+1}[/TEX]
[TEX]= \sqrt[]{x}-1+\frac{9}{\sqrt[]{x}+1}[/TEX]
[TEX]= \sqrt[]{x}+1+\frac{9}{\sqrt[]{x}+1}-2[/TEX]
[TEX]\geq 6-2=4[/TEX]
Vậy Min [TEX]\frac{x+8}{\sqrt[]{x}+1} = 4[/TEX] khi x=4
 
H

huynhbachkhoa23

Cách khác:

$a=\sqrt{x} \ge 0$

$A=\dfrac{a^2+8}{a+1}=m >0$

$\rightarrow a^2-ma-m+8=0$

$\Delta = m^2+4m-32=(m+8)(m-4) \ge 0 \leftrightarrow m \ge 4$

$\text{minA}=4 \leftrightarrow x=4$
 
You must log in or register to reply here.
Top Bottom