[Toán 9]­Tìm min

D

dotantai1999

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E

eye_smile

Biến đổi BT trở thành:
$A=\dfrac{\sqrt{x-1}+2}{\sqrt{x-1}+3}=1-\dfrac{1}{\sqrt{x-1}+3}$
BT có GTNN \Leftrightarrow $\dfrac{1}{\sqrt{x-1}+3}$ lớn nhất
\Leftrightarrow $\sqrt{x-1}=0$
\Leftrightarrow x=1


Biến đổi:
$A=\dfrac{x+3\sqrt{x-1}+1}{x+4\sqrt{x-1}+2}=\dfrac{x-1+3\sqrt{x-1}+2}{x-1+4\sqrt{x-1}+3}=\dfrac{(\sqrt{x-1}+1)(\sqrt{x-1}+2)}{(\sqrt{x-1}+1)(\sqrt{x-1}+3)}=\dfrac{\sqrt{x-1}+2}{\sqrt{x-1}+3}$
 
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