{(14−3x)√4−x=(3y+11)√3+y(1)3√x+6+2√3−y=6(2) (1)⇔2√4−x+3(4−x)√4−x=2√3+y+3(3+y)√3+y⇒4−x=3+y(2)⇔3√x+6+2√x+2=6$\Leftrightarrow x=2 \vee \frac{1}{\sqrt[3]{x+6}^{2}+2\sqrt[3]{x+6}+4}=\frac{1}{\sqrt{x+2}+2}$ (vô nghiệm)
{(14−3x)√4−x=(3y+11)√3+y(1)3√x+6+2√3−y=6(2) (1)⇔2√4−x+3(4−x)√4−x=2√3+y+3(3+y)√3+y⇒4−x=3+y(2)⇔3√x+6+2√x+2=6$\Leftrightarrow x=2 \vee \frac{1}{\sqrt[3]{x+6}^{2}+2\sqrt[3]{x+6}+4}=\frac{2}{\sqrt{x+2}+2}$ (vô nghiệm)