Question

Se consideră numerele a=

$\frac{4}{ \sqrt{6 + 2 \sqrt{5} } }$
si b=
$\sqrt{5} - 1$
​

Answer 1

 

$\displaystyle\bf\\a=\frac{4}{\sqrt{6+2\sqrt{5}}}\\\\\\a=\frac{4}{\sqrt{5+1+2\sqrt{5}}}\\\\\\a=\frac{4}{\sqrt{5+2\sqrt{5}+1}}\\\\\\a=\frac{4}{\sqrt{\left(\sqrt{5}\right)^2+2\times\sqrt{5}\times1+1^{^2}}}\\\\\\a=\frac{4}{\sqrt{\left(\sqrt{5}+1\right)^2}}\\\\\\a=\frac{4}{\sqrt{5}+1}\\\\\\a=\frac{4(\sqrt{5}-1)}{(\sqrt{5}+1)(\sqrt{5}-1)}=\frac{4(\sqrt{5}-1)}{5-1} =\frac{4(\sqrt{5}-1)}{4}\\\\\\\boxed{\bf~a=\sqrt{5}-1}\\\\\\\boxed{\bf~b=\sqrt{5}-1}\\\\\\a=b\implies \boxed{\bf~m_g=a=b=\sqrt{5}-1}}$