[Đại 9]cho biểu thức

H

hocsinhchankinh

Đặt $t=\sqrt{x-1} (t \ge 0)$

BT trở thành:

$A=\dfrac{t^2-9}{t-3} \Longleftrightarrow A=t+3 \ge 3$

Dấu "=" xảy ra $\Longleftrightarrow t=0 \Longleftrightarrow x=1$
 
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L

lp_qt

$A=\dfrac{x-10}{\sqrt{x-1}-3}=\dfrac{3(\sqrt{x-1}-3)+(x-1-3\sqrt{x-1})}{\sqrt{x-1}-3}=3+\dfrac{(\sqrt{x-1}-3).\sqrt{x-1}}{\sqrt{x-1}-3}=3+\sqrt{x-1} \ge 3$
 
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