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Einstein55
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Răspuns :

[tex]\sqrt{7+4\sqrt3}=\sqrt{4+3+4\sqrt3}=\sqrt{2^2+2\cdot2\sqrt3+(\sqrt3)^2}=\sqrt{(2+\sqrt3)^2}=2+\sqrt3[/tex]

Proporția devine:

[tex]\it \dfrac{x}{10}=\dfrac{2+\sqrt3}{2+\sqrt3} \Rightarrow \dfrac{x}{10}=1 \Rightarrow x=10[/tex]

 

[tex]\displaystyle\bf\\\frac{x}{10}=\frac{2+\sqrt{3}}{\sqrt{7+4\sqrt{3}}}\\\\Ne~ocupam~de~radicalul~compus~de~la~numitor\\\\\frac{x}{10}=\frac{2+\sqrt{3}}{\sqrt{7+2\cdot2\cdot\sqrt{3}}} ~~~Observam~ca~2^2+\Big(\sqrt{3}\Big)^2=7\\\\\\\frac{x}{10}=\frac{2+\sqrt{3}}{\sqrt{4+3+2\cdot2\cdot\sqrt{3}}}[/tex]

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[tex]\displaystyle\bf\\\frac{x}{10}=\frac{2+\sqrt{3}}{\sqrt{4+2\cdot2\cdot\sqrt{3}+3}}\\\\\\\frac{x}{10}=\frac{2+\sqrt{3}}{\sqrt{2^2+2\cdot2\cdot\sqrt{3}+\Big(\sqrt{3}\Big)^2}}\\\\\\\frac{x}{10}=\frac{2+\sqrt{3}}{\sqrt{\Big(2+\sqrt{3}\Big)^2}}\\\\\\\frac{x}{10}=\frac{2+\sqrt{3}}{2+\sqrt{3}}\\\\\\\frac{x}{10}=1\\\\\\\implies~~\boxed{\bf x=10}[/tex]

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 10 ∈ N si N ⊂ Z ⊂ R

⇒ Orice numar natural este si numar real.