Question

Va rog problema 10
Dau coroana

Answer 1

$\displaystyle\\a=\sqrt{4-\sqrt{15}} + \sqrt{4+\sqrt{15}}\\\\\text{Avem radicali compusi.}\\\text{Folosim formula:}\\\\\sqrt{a\pm\sqrt{b}}=\sqrt{\frac{a+c}{2}}\pm\sqrt{\frac{a-c}{2}}~~~~~\text{unde } ~c = \sqrt{a^2-b}$

$\displaystyle\\\text{Rezolvare:}\\\\c = \sqrt{4^2 - 15}=\sqrt{16 - 15}=\sqrt{1}=1\\\\\sqrt{4-\sqrt{15}} = \sqrt{\frac{4+1}{2}}-\sqrt{\frac{4-1}{2}}=\boxed{\sqrt{\frac{5}{2}}-\sqrt{\frac{3}{2}}}\\\\\sqrt{4+\sqrt{15}} = \sqrt{\frac{4+1}{2}}+\sqrt{\frac{4-1}{2}}=\boxed{\sqrt{\frac{5}{2}}+\sqrt{\frac{3}{2}}}$

$\displaystyle\\a=\sqrt{4-\sqrt{15}} + \sqrt{4+\sqrt{15}} =\\\\= \sqrt{\frac{5}{2}}-\sqrt{\frac{3}{2}}+ \sqrt{\frac{5}{2}}+\sqrt{\frac{3}{2}} =\sqrt{\frac{5}{2}}+\sqrt{\frac{5}{2}}=2\sqrt{\frac{5}{2}}=\sqrt{\frac{4\cdot5}{2}}=\boxed{\sqrt{10} }\\\\a = \sqrt{10} \\\\\Longrightarrow~~~a^2 = (\sqrt{10})^2 = \boxed{\bf 10}$

Raspuns corect B