Ta thấy :
\(\sqrt{x^2-4x+5}=\sqrt{\left(x^2-4x+4\right)+1}=\sqrt{\left(x-2\right)^2+1}\ge\sqrt{1}=1\)
\(\sqrt{x^2-4x+8}=\sqrt{\left(x^2-4x+4\right)+4}=\sqrt{\left(x-2\right)^2+4}\ge\sqrt{4}=2\)
\(\sqrt{x^2-4x+9}=\sqrt{\left(x^2-4x+4\right)+5}=\sqrt{\left(x-2\right)^2+5}\ge\sqrt{5}\)
\(\Rightarrow\sqrt{x^2-4x+5}+\sqrt{x^2-4x+8}+\sqrt{x^2-4x+9}\ge3+\sqrt{5}\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}\sqrt{\left(x-2\right)^2+1}=1\\\sqrt{\left(x-2\right)^2+4}=2\\\sqrt{\left(x-2\right)^2+5}=\sqrt{5}\end{cases}\Rightarrow x=2}\)
Vậy \(x=2\)
\(\sqrt{x^2-4x+5}+\sqrt{x^2-4x+8}+\sqrt{x^2-4x+9}=3+\sqrt{5}\)
\(\Leftrightarrow\sqrt{\left(x-2\right)^2+1}+\sqrt{\left(x-2\right)^2+4}+\sqrt{\left(x-2\right)^2+5}=3+\sqrt{5}\)
Vế trái \(T\ge\sqrt{1}+\sqrt{4}+\sqrt{5}=3+\sqrt{5}\)
Dấu "=" xảy ra khi và chỉ khi (x-2)2=0 <=> x=2
