Question

Aratati ca numerele a,b,c sunt intregi:
a=4√10-2√8-(6+4√10)=
b=2(4+√21)-6√7+2√7×(3-√3)=
c=4√3(√8-√3) -2√2 × (3√3-√2)=

Answer 1

$a=4 \sqrt{10} -2 \sqrt{8} -(6+4 \sqrt{10} ) \\ a=4 \sqrt{10} -2*2 \sqrt{2} -6-4 \sqrt{10} \\ a=4 \sqrt{10} -4 \sqrt{2} -6-4 \sqrt{10} \\ a=-4 \sqrt{2} -6 \\ b=2(4+ \sqrt{21} )-6 \sqrt{7} +2 \sqrt{7} (3- \sqrt{3} ) \\ b=8+2 \sqrt{21} -6 \sqrt{7} +6 \sqrt{7} -2 \sqrt{21} \\ b=8 \\ c=4 \sqrt{3} ( \sqrt{8} - \sqrt{3} )-2 \sqrt{2} (3 \sqrt{3} - \sqrt{2} ) \\ c=4 \sqrt{3} (2 \sqrt{2} - \sqrt{3} )-6 \sqrt{6} +4 \\ c=8 \sqrt{6} -12-6 \sqrt{6} +4 \\ c=2 \sqrt{6} -8$