Question

ajutați ma va rog ofer coroană​

Answer 1

$a=\left(\sqrt{7}-\sqrt{5}\right)^{-1}$

stim ca $a^{-1}=\frac{1}{a}$

$a=\frac{1}{\sqrt{7}-\sqrt{5}}$

$\frac{^{\sqrt{7}+\sqrt{5})}1}{\sqrt{7}-\sqrt{5}} =\frac{1\cdot \left(\sqrt{7}+\sqrt{5}\right)}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)} =\frac{\sqrt{7}+\sqrt{5}}{2}$

$b=\left(\sqrt{5}-\sqrt{3}\right)^{-1}$

$b=\frac{1}{\sqrt{5}-\sqrt{3}}$

$\frac{^{\sqrt{5}+\sqrt{3})}1}{\sqrt{5}-\sqrt{3}} =\frac{1\cdot \left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)} =\frac{\sqrt{5}+\sqrt{3}}{2}$

$\frac{\sqrt{7}+\sqrt{5}}{2} > \frac{\sqrt{5}+\sqrt{3}}{2}$

$a > b$