ĐK: \(x\ge\dfrac{9}{5}\)
\(K=\sqrt{5x+6\sqrt{5x-9}}+\sqrt{5x-6\sqrt{5x-9}}\\ =\sqrt{\left(5x-9\right)+6\sqrt{5x-9}+9}+\sqrt{\left(5x-9\right)-6\sqrt{5x-9}+9}\\ =\sqrt{\left(\sqrt{5x-9}+3\right)^2}+\sqrt{\left(\sqrt{5x-9}-3\right)^2}\\ =\left|3+\sqrt{5x-9}\right|+\left|3-\sqrt{5x-9}\right|\\ \ge\left|3+\sqrt{5x-9}+3-\sqrt{5x-9}\right|\\ =6\)
\(\Rightarrow K\ge6\)
Dấu \(=\) xảy ra \(\Leftrightarrow\left(3+\sqrt{5x-9}\right)\left(3-\sqrt{5x-9}\right)\ge0\)
\(\Leftrightarrow9-\left(5x-9\right)\ge0\Leftrightarrow18-5x\ge0\Leftrightarrow x\le\dfrac{18}{5}\)
Kết hợp với điều kiện ta được: \(K_{min}=6\) đạt được khi \(\dfrac{9}{5}\le x\le\dfrac{18}{5}\).
