Question

\(A=\sqrt{m^2+2m+1}+\sqrt{m^2-2m+1}.A_{min}=?\)

Answer 1

Lời giải:
$A=\sqrt{(m+1)^2}+\sqrt{(m-1)^2}=|m+1|+|m-1|$

$=|m+1|+|1-m|\geq |m+1+1-m|=2$

Vậy $A_{\min}=2$. Giá trị này đạt tại $(m+1)(1-m)\geq 0$

$\Leftrightarrow -1\leq m\leq 1$

Answer 2

\(A=\sqrt{m^2+2m+1}+\sqrt{m^2-2m+1}=\sqrt{\left(m+1\right)^2}+\sqrt{\left(m-1\right)^2}=\left|m+1\right|+\left|m-1\right|=\left|m+1\right|+\left|1-m\right|\ge\left|m+1+1-m\right|=2\)

\(minA=2\Leftrightarrow\left(m+1\right)\left(1-m\right)\ge0\Leftrightarrow1-m^2\ge0\Leftrightarrow m^2\le1\Leftrightarrow-1\le m\le1\)