Đáp án: $x=0$
Giải thích các bước giải:
Ta có:
$P=\sqrt{x^2+x+1}+\sqrt{x^2-x+1}$
$\to P=\sqrt{(x+\dfrac12)^2+(\dfrac{\sqrt{3}}2)^2}+\sqrt{(x-\dfrac12)^2+(\dfrac{\sqrt{3}}2)^2}$
$\to P=\sqrt{(x+\dfrac12)^2+(\dfrac{\sqrt{3}}2)^2}+\sqrt{(\dfrac12-x)^2+(\dfrac{\sqrt{3}}2)^2}$
$\to P\ge \sqrt{(x+\dfrac12+\dfrac12-x)^2+(\dfrac{\sqrt{3}}2+\dfrac{\sqrt{3}}2)^2}$
$\to P\ge 2$
Mà $\sqrt{x^2+x+1}+\sqrt{x^2-x+1}=2$
$\to$Dấu = xảy ra khi:
$\dfrac{x+\dfrac12}{\dfrac12-x}=\dfrac{\dfrac{\sqrt{3}}2}{\dfrac{\sqrt{3}}2}$
$\to x=0$
