Question

Aratati ca numarul n=(radical 6- radical3)^2 +( radical 2 +3)^2 + ( radical 3-radical 6) • (radical 6+radical 3) este un numar natural

Answer 1

$n=(\sqrt{6}-3)^{2} +(\sqrt{2}+3)^{2}+(\sqrt{3}-\sqrt{6})(\sqrt{6}+\sqrt{3})=$

$n=\sqrt{6}^{2}-6\sqrt{6}+3^{2} +\sqrt{2}^{2}+6\sqrt{2}+3^{2}+(\sqrt{3}-\sqrt{6})(\sqrt{3}+\sqrt{6})$

$n=6+6\sqrt{6}+9+2+6\sqrt{6}+9+(3-6)$

$n=6+9+2+9+(-3)$

$n=26-3$

$\boxed{n=23}$

n $\in$ |N

Formule de calcul prescurtat

$\boxed{ (a+b)^{2}=a^{2}+2ab+b^{2}}$

$\boxed{ (a-b)^{2}=a^{2}-2ab+b^{2}}$

$\boxed{ (a+b)(a-b)=a^{2}-b^{2}}$