[imath]\displaystyle \sqrt{7+4 \sqrt{3}} + \sqrt{2} \sqrt{14-5 \sqrt{3}}[/imath]
[imath]= \displaystyle \sqrt{2^2+2. \sqrt{3}.2 + ( \sqrt{3})^2} + \sqrt{2(14-5 \sqrt{3})} \\
= \sqrt{(2+ \sqrt{3})^2} + \sqrt{28-10 \sqrt{3}} \\
= \sqrt{(2+ \sqrt{3})^2} + \sqrt{5^2-2. 5. \sqrt{3} + ( \sqrt{3})^2} \\
= \sqrt{(2+ \sqrt{3})^2} + \sqrt{(5- \sqrt{3})^2} \\
=|2+ \sqrt{3}| + |5- \sqrt{3}| \\
=2+ \sqrt{3} + 5- \sqrt{3} \ \ (vi \ 5 > \sqrt{3}) \\
=7[/imath]