Đáp án: $ P\ge \sqrt{2}\sqrt{5+\sqrt{22}}$
Giải thích các bước giải:
Ta có:
$P=\sqrt{2x^2-6x+10}+\sqrt{2x^2-8x+12}$
$\to P=\sqrt{2}(\sqrt{x^2-3x+5}+\sqrt{x^2-4x+6})$
$\to P=\sqrt{2}(\sqrt{\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}}+\sqrt{(x-2)^2+2})$
$\to P=\sqrt{2}(\sqrt{\left(x-\dfrac{3}{2}\right)^2+\sqrt{\dfrac{11}{4}}^2}+\sqrt{(2-x)^2+(\sqrt{2})^2})$
$\to P\ge \sqrt{2}\sqrt{\left(x-\dfrac{3}{2}+2-x\right)^2+(\sqrt{\dfrac{11}{4}}+\sqrt{2})^2}$
$\to P\ge \sqrt{2}\sqrt{5+\sqrt{22}}$
Dấu = xảy ra khi:
$\dfrac{x-\dfrac32}{2-x}=\dfrac{\sqrt{\dfrac{11}{4}}}{\sqrt{2}}$
$\to x=\dfrac{10-\sqrt{22}}{3}$
