\(=\sqrt{3\left(x^2-2x+1\right)+25}\supseteq\sqrt{3\left(x+1\right)^2+25}\supseteq5\)
min=5 <=>x=-1
\(\text{Đặt }A=\sqrt{3x^2-6x+28}=\sqrt{3x^2-6x+3+25}\)
\(=\sqrt{3.\left(x^2-2x+1\right)+25}=\sqrt{3.\left(x-1\right)^2+25}\)
\(\Rightarrow A^2=3.\left(x-1\right)^2+25\ge25\Rightarrow A\ge\sqrt{25}=5\)
Dấu "=" xảy ra khi : x=1
Vậy GTNN của A là 5 tại x=1
\(=\sqrt{3}\left(\sqrt{x^2-2x+\frac{28}{3}}\right)\)
\(=\sqrt{3}\left(\sqrt{\left(x-1\right)^2+\frac{25}{3}}\right)\)
\(=\sqrt{3\left(x-1\right)^2}+\sqrt{3}\sqrt{\frac{25}{3}}\)
\(=\sqrt{3}\left(x-1\right)+5\ge5\)
Vậy Min đề = 5 khi x - 1 = 0 => x = 1
