Question

Calculati:

(3-2√2)$\sqrt{17+12 \sqrt{2} }$ +$\sqrt{7+4 \sqrt{3} }$ (√3-2)

Answer 1

     $(3-2 \sqrt{2}) \sqrt{17+12 \sqrt{2}} +\sqrt{7+4\sqrt{3}}( \sqrt{3}-2)= \\ =(3-2 \sqrt{2}) \sqrt{9+12 \sqrt{2}+8} +\sqrt{3+4\sqrt{3}+4}( \sqrt{3}-2)= \\ =(3-2 \sqrt{2}) \sqrt{3^{2}+2*3*2\sqrt{2}+ (2\sqrt{2})^{2}} +\sqrt{3+4\sqrt{3}+2^{2}}( \sqrt{3}-2)= \\=(3-2 \sqrt{2}) \sqrt{(3+2\sqrt{2})^{2}} +\sqrt{(\sqrt{3}+2)^{2}}( \sqrt{3}-2)= \\ =(3-2 \sqrt{2}) (3+2\sqrt{2}) +(\sqrt{3}+2)(\sqrt{3}-2)= \\=(3^{2}-(2\sqrt{2})^{2}) +((\sqrt{3})^{2}-(2)^{2})= \\ =(9-8)+(3-4)= \\ =(1)+(-1)= \\ =1-1=0$