[Toán 9] Căn bậc hai

E

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$\sqrt{{x^2}-4x+5}=\sqrt{{(x-2)^2}+1}$ \geq $\sqrt{1}=1$

$\sqrt{{x^2}-4x+8}=\sqrt{{(x-2)^2}+4}$ \geq $\sqrt{4}=2$

$\sqrt{{x^2}-4x+9}=\sqrt{{(x-2)^2}+5}$ \geq $\sqrt{5}$

\Rightarrow VT \geq VP

Dấu"=" xảy ra \Leftrightarrow x=2
 
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