[Toán 9] Tìm giá trị nhỏ nhất.

Q

quanghuy17111999@gmail.com

\sqrt{x}-2 /\sqrt{x} +1 =\sqrt{x}+1-3 /\sqrt{x}+1=1-3/\sqrt{x}+1





Đến đây chắc bạn biết giải rồi chúc may mắn:)
 
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B

brainy1610

[TEX]A=\frac{\sqrt{x}-2}{\sqrt{x} +1}[/TEX]
[TEX]= \frac{\sqrt{x}+1-3}{\sqrt{x}+1}[/TEX]
[TEX]= \frac{\sqrt{x}+1}{\sqrt{x}+1}+\frac{-3}{\sqrt{x}+1}[/TEX]
[TEX]=1+\frac{-3}{\sqrt{x}+1}[/TEX]
=> A MIN=1 khi x=(-1)
 
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C

congchuaanhsang

ĐKXĐ x\geq0

A=$\dfrac{\sqrt{x}-2}{\sqrt{x}+1}$=$1-\dfrac{3}{\sqrt{x}+1}$

$\sqrt{x}+1$\geq1\Rightarrow$\dfrac{3}{\sqrt{x}+1}$\leq3

\Rightarrow$1-\dfrac{3}{\sqrt{x}+1}$\geq-2

$A_{min}$=-2\Leftrightarrowx=0
 
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