Áp dụng BĐT Cô si ta có \(x+\dfrac{4}{x}\ge2\sqrt{x.\dfrac{4}{x}}=2\sqrt{4}=4\)

Dấu = xảy ra \(\Leftrightarrow x=\dfrac{4}{x}\Leftrightarrow x^2=4\Leftrightarrow x=2\) (Vì \(1\le x\le3\))

=> (2x + 3)³ = (-3)³

=> 2x + 3 = -3

=> 2x = -6

=> x = -3

Vậy x = -3

\(\left(\dfrac{2}{\sqrt{3}-1}+\dfrac{3}{\sqrt{3}-2}+\dfrac{15}{3-\sqrt{3}}\right).\dfrac{1}{\sqrt{3}+5}\\ =\left(\dfrac{2.\sqrt{3}\left(\sqrt{3}-2\right)}{\sqrt{3}\left(\sqrt{3}-1\right)\left(\sqrt{3}-2\right)}+\dfrac{3\sqrt{3}\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-2\right)\sqrt{3}\left(\sqrt{3}-1\right)}+\dfrac{15\left(\sqrt{3}-2\right)}{\left(\sqrt{3}-2\right)\sqrt{3}\left(\sqrt{3}-1\right)}\right).\dfrac{1}{\sqrt{3}+5}\\ =\dfrac{-15+8\sqrt{3}}{\left(\sqrt{3}-2\right)\sqrt{3}\left(\sqrt{3}-1\right)}.\dfrac{1}{\sqrt{3}+5}\\ =\dfrac{\sqrt{3}\left(8-5\sqrt{3}\right)}{\left(\sqrt{3}-2\right)\sqrt{3}\left(\sqrt{3}-1\right)}.\dfrac{1}{\sqrt{3}+5}\\ =\dfrac{8-5\sqrt{3}}{\left(5-3\sqrt{3}\right)\left(\sqrt{3}+5\right)}=\dfrac{8-5\sqrt{3}}{16-10\sqrt{3}}=\dfrac{1}{2}\)