Question

Aratati ca numarul x este natural, unde:
a) x= totul sub radical 11+6√2 - totul sub radical 9-4√2 - totul sub radical 6-4√2
b)x= totul sub radical 12+6√3 - 2 totul sub radical 7-4√3 -3 totul sub radical 4-2√3

Answer 1

[tex]a)x=\sqrt{11+6\sqrt2}}-\sqrt{9-4\sqrt2}}-\sqrt{6-4\sqrt2}\\ Luam\ fiecare\ radical\ in\ parte\ si\ il\ rezolvam:\\ \sqrt{11+6\sqrt2}=\sqrt{9+2+6\sqrt2}=\sqrt{(3+\sqrt2)^2}=3+\sqrt2\\ \sqrt{9-4\sqrt2}=\sqrt{8+1-4\sqrt2}=\sqrt{(2\sqrt2-1)^2}=2\sqrt2-1\\ \sqrt{6-4\sqrt2}=\sqrt{4+2-4\sqrt2}=\sqrt{(2-\sqrt2)^2}=2-\sqrt2\\ Revenind:\\ x=3+\sqrt2-2\sqrt2+1-2+\sqrt2=2\in N\\ [/tex]
[tex]b)x=\sqrt{12+6\sqrt3}-2\sqrt{7-4\sqrt3}-3\sqrt{4-2\sqrt3}}\\ Facem\ la\ fel\ ca\ la\ a:\\ \sqrt{12+6\sqrt3}=\sqrt{9+3+6\sqrt3}=\sqrt{(3+\sqrt3)^2}=3+\sqrt3\\ \sqrt{7-4\sqrt3}=\sqrt{4+3-4\sqrt3}=\sqrt{(2-\sqrt3)^2}=2-\sqrt3\\ \sqrt{4-2\sqrt3}=\sqrt{3+1-2\sqrt3}=\sqrt{(\sqrt3-1)^2}=\sqrt3-1\\ Revenind:\\ x=3+\sqrt3-2(2-\sqrt3)-3(\sqrt3-1)\\ x=3+\sqrt3-4+2\sqrt3-3\sqrt3+3\\ x=2 \in N[/tex]