[toán 9] Đại số

I

iceghost

$A=\sqrt{x^2 + 4x + 8}=\sqrt{x^2 + 4x + 4+4}=\sqrt{(x+2)^2+4}\ge \sqrt4=2 \\
\implies Min_A=2 \iff x+2=0 \iff x=-2$
 
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