Question

Dauu coroanaa
Help!!!!!!!
Ex 17 ......20 puncteee

Answer 1

$A=\frac{\sqrt{10+2*\sqrt{22+6*\sqrt{4+2\sqrt{3}}}}}{\sqrt{10-2*\sqrt{22-6*\sqrt{4-2\sqrt{3}}}}}\\\\\\A_1={\sqrt{10+2*\sqrt{22+6*\sqrt{4+2\sqrt{3}}}}}\\\\A_1={\sqrt{10+2*\sqrt{22+6*\sqrt{3+2\sqrt{3}+1}}}}\\\\A_1={\sqrt{10+2*\sqrt{22+6*\sqrt{(\sqrt3+1)^2}}}}\\\\A_1={\sqrt{10+2*\sqrt{22+6*|\sqrt3+1|}}}}\\\\A_1={\sqrt{10+2*\sqrt{22+6*(\sqrt3+1)}}}\\\\A_1={\sqrt{10+2*\sqrt{22+6\sqrt3+6)}}}\\\\A_1={\sqrt{10+2*\sqrt{27+6*\sqrt3+1)}}}\\\\A_1={\sqrt{10+2*\sqrt{(3\sqrt3+1)^2}}}$

$A_1={\sqrt{10+2*|3\sqrt3+1|}}\\\\A_1={\sqrt{10+2*(3\sqrt3+1)}}\\\\A_1={\sqrt{10+6\sqrt3+2}}\\\\A_1={\sqrt{10+6\sqrt3+2}}\\\\A_1={\sqrt{9+6\sqrt3+3}}\\\\A_1={\sqrt{(3+\sqrt3)^2}}\\\\A_1=|3+\sqrt3|=3+\sqrt3$

$A_2={\sqrt{10-2*\sqrt{22-6*\sqrt{4-2\sqrt{3}}}}}\\\\A_2={\sqrt{10-2*\sqrt{22-6*\sqrt{3-2\sqrt{3}+1}}}}\\\\A_2={\sqrt{10-2*\sqrt{22-6*\sqrt{(\sqrt3-1)^2}}}}\\\\A_2={\sqrt{10-2*\sqrt{22-6*|\sqrt3-1|}}}}\\\\A_2={\sqrt{10-2*\sqrt{22-6*(\sqrt3-1)}}}\\\\A_2={\sqrt{10-2*\sqrt{22-6\sqrt3+6)}}}\\\\A_2={\sqrt{10-2*\sqrt{27-6*\sqrt3+1)}}}\\\\A_2={\sqrt{10-2*\sqrt{(3\sqrt3-1)^2}}}$

$A_2={\sqrt{10-2*|3\sqrt3-1|}}\\\\A_2={\sqrt{10-2*(3\sqrt3-1)}}\\\\A_2={\sqrt{10-6\sqrt3+2}}\\\\A_2={\sqrt{10-6\sqrt3+2}}\\\\A_2={\sqrt{9-6\sqrt3+3}}\\\\A_2={\sqrt{(3-\sqrt3)^2}}\\\\A_2=|3-\sqrt3|=3-\sqrt3\\\\\\A=\frac{A_1}{A_2}=\frac{3+\sqrt3}{3-\sqrt3}=\frac{(3+\sqrt3)(3+\sqrt3)}{(3+\sqrt3)(3-\sqrt3)}\\\\A=\frac{3^2+2*3*\sqrt3+\sqrt3^2}{3^2-\sqrt3^2}=\frac{9+6\sqrt3+3}{9-3}=\frac{12+6\sqrt3}{6}=\frac{6(2+\sqrt3)}{6}=2+\sqrt3$

Observatii: Poti folosi si metoda radicalilor compusi, eu insa am descompus in termeni pentru a putea aplica formulele

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

P.S= Sper ca nu am gresit la calcule, daca da imi cer scuze.