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aratai ca x=√9+4√2(totul sub radical)+√(3-√2)²(totul sub radical)-√3-2√2(totul sub radical)este numar natura;

Răspuns :

[tex]\sqrt{9+4\sqrt2}=\sqrt{(2\sqrt2)^2+2\cdot2\sqrt2\cdot1+1}=\sqrt{(2\sqrt2+1)^2}=2\sqrt2+1\\ \sqrt{(3-\sqrt2)^2}=3-\sqrt2\\ \sqrt{3-2\sqrt2}=\sqrt{\sqrt2^2-2\sqrt2\cdot1+1^2}=\sqrt{(\sqrt2-1)^2}=\sqrt2-1\\ 2\sqrt2+1+3-\sqrt2-(\sqrt2-1)=2\sqrt2+1+3-\sqrt2-\sqrt2+1=4\in\mathbb{N}[/tex]
 √9 + 4√2 = √ (2√2)² + 2·2√2·1 + 1² = √( 2√2+1) ² = 2√2 + 1 
√( 3 - √2)² = I 3- √2I = pozitiv  = 3 - √2
√3 -2√2   = √(√2)² - 2√2 +1² = √(√2-1)² = I√2-1I = pozitiv = √2 -1 
x = 2√2 + 1 + 3 - √2 -√2  + 1 = 5 ∈ N