Question

va rog sa ma ajutați la exercițiul 12 și 13 îmi trebuie urgent... va rog frumos....

Answer 1

12.

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

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

$\frac{\sqrt{2}}{\sqrt{3}-2\sqrt{5}}=\frac{\sqrt{2}\left(\sqrt{3}+2\sqrt{5}\right)}{\left(\sqrt{3}-2\sqrt{5}\right)\left(\sqrt{3}+2\sqrt{5}\right)}=\frac{\sqrt{2}\left(\sqrt{3}+2\sqrt{5}\right)}{-17}=-\frac{\sqrt{2}\left(2\sqrt{5}+\sqrt{3}\right)}{17}$

13.

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

$\frac{\sqrt{2}-1}{\sqrt{3}+2}+\frac{\sqrt{2}+1}{\sqrt{3}-2}=\frac{\left(\sqrt{2}-1\right)\left(\sqrt{3}-2\right)}{\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)}+\frac{\left(\sqrt{2}+1\right)\left(2+\sqrt{3}\right)}{\left(\sqrt{3}-2\right)\left(2+\sqrt{3}\right)}=\frac{\left(\sqrt{2}-1\right)\left(\sqrt{3}-2\right)+\left(\sqrt{2}+1\right)\left(2+\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(\sqrt{3}-2\right)}=\frac{4+2\sqrt{6}}{-1}=-\left(4+2\sqrt{6}\right)=-4-2\sqrt{6}$

$\frac{6-2\sqrt{5}}{\sqrt{5}-1}+\frac{6+2\sqrt{5}}{\sqrt{5}+1}=\frac{\left(6-2\sqrt{5}\right)\left(1+\sqrt{5}\right)}{\left(\sqrt{5}-1\right)\left(1+\sqrt{5}\right)}+\frac{\left(6+2\sqrt{5}\right)\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}=\frac{\left(6-2\sqrt{5}\right)\left(1+\sqrt{5}\right)+\left(6+2\sqrt{5}\right)\left(\sqrt{5}-1\right)}{\left(\sqrt{5}-1\right)\left(1+\sqrt{5}\right)}=\frac{8\sqrt{5}}{4}=2\sqrt{5}$