Question

Sa se arate că nr. b=(2-radical2)^2 + radical (radical3 +2rad2) ^2+rad( 11-4rad 6)

Answer 1

b=$(2- \sqrt{2})^2+ \sqrt{( \sqrt{3}+2 \sqrt{2})^2}+ \sqrt{11-4 \sqrt{6} }=4-4 \sqrt{2}+2+ \sqrt{3}+2 \sqrt{2}+ \sqrt{(2 \sqrt{2}- \sqrt{3})^2}$=6-2$\sqrt{2}+ \sqrt{3}+2 \sqrt{2}- \sqrt{3}$=6