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Ta có BĐT sau:
$\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\ge\sqrt{(a+c)^2+(b+d)^2}$
Dấu bằng xảy ra khi: $ad=bc$
Áp dụng BĐT trên ta có:
$y=\sqrt{(\frac{1}{2}-x)^2+(\frac{\sqrt3}{2})^2}+\sqrt{(x+\frac{1}{2})^2+(\frac{\sqrt3}{2})^2}$
$\ge\sqrt{(\frac{1}{2}-x+x+\frac{1}{2})^2+(\frac{\sqrt3}{2}+\frac{\sqrt3}{2})^2}=2$
Min$y=2\Leftrightarrow \frac{\sqrt3}{2}(\frac{1}{2}-x)=\frac{\sqrt3}{2}(x+\frac{1}{2})\Leftrightarrow x=0$
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