Répondre :
Bonsoir,
[tex]a)\ \dfrac{1}{3+\sqrt{2}}=\dfrac{3-\sqrt{2}}{(3+\sqrt{2})(3-\sqrt{2})}=\dfrac{3-\sqrt{2}}{9-2}\\\\\\=\dfrac{3-\sqrt{2}}{7}=\dfrac{1}{7}(3-\sqrt{2})\\\\\\b)\ \dfrac{2}{\sqrt{3}-\sqrt{5}}=\dfrac{2(\sqrt{3}+\sqrt{5})}{(\sqrt{3}-\sqrt{5})(\sqrt{3}+\sqrt{5})}=\dfrac{2(\sqrt{3}+\sqrt{5})}{3-5}\\\\\\=\dfrac{2(\sqrt{3}+\sqrt{5})}{-2}=-(\sqrt{3}+\sqrt{5})[/tex]
[tex]c)\ \dfrac{\sqrt{3}}{2\sqrt{3}-1}=\dfrac{\sqrt{3}(2\sqrt{3}+1)}{(2\sqrt{3}-1)(2\sqrt{3}+1)}=\dfrac{2\times3+\sqrt{3}}{12-1}\\\\\\=\dfrac{6+\sqrt{3}}{11}=\dfrac{1}{11}(6+\sqrt{3})\\\\\\d)\ \dfrac{3\sqrt{2}}{\sqrt{2}+2\sqrt{3}}=\dfrac{3\sqrt{2}(\sqrt{2}-2\sqrt{3})}{(\sqrt{2}+2\sqrt{3})(\sqrt{2}-2\sqrt{3})}=\dfrac{3\times2-6\sqrt{6}}{2-12}\\\\\\=\dfrac{6-6\sqrt{6}}{-10}=\dfrac{6(1-\sqrt{6})}{-10}=-\dfrac{3}{5}(1-\sqrt{6})[/tex]
[tex]e)\ \dfrac{2\sqrt{3}}{2\sqrt{3}-5\sqrt{2}}=\dfrac{2\sqrt{3}(2\sqrt{3}+5\sqrt{2})}{(2\sqrt{3}-5\sqrt{2})(2\sqrt{3}+5\sqrt{2})}=\dfrac{4\times3+10\sqrt{6}}{12-50}\\\\\\=\dfrac{12+10\sqrt{6}}{-38}=\dfrac{2(6+5\sqrt{6})}{-38}=-\dfrac{1}{19}(6+5\sqrt{6})[/tex]
[tex]a)\ \dfrac{1}{3+\sqrt{2}}=\dfrac{3-\sqrt{2}}{(3+\sqrt{2})(3-\sqrt{2})}=\dfrac{3-\sqrt{2}}{9-2}\\\\\\=\dfrac{3-\sqrt{2}}{7}=\dfrac{1}{7}(3-\sqrt{2})\\\\\\b)\ \dfrac{2}{\sqrt{3}-\sqrt{5}}=\dfrac{2(\sqrt{3}+\sqrt{5})}{(\sqrt{3}-\sqrt{5})(\sqrt{3}+\sqrt{5})}=\dfrac{2(\sqrt{3}+\sqrt{5})}{3-5}\\\\\\=\dfrac{2(\sqrt{3}+\sqrt{5})}{-2}=-(\sqrt{3}+\sqrt{5})[/tex]
[tex]c)\ \dfrac{\sqrt{3}}{2\sqrt{3}-1}=\dfrac{\sqrt{3}(2\sqrt{3}+1)}{(2\sqrt{3}-1)(2\sqrt{3}+1)}=\dfrac{2\times3+\sqrt{3}}{12-1}\\\\\\=\dfrac{6+\sqrt{3}}{11}=\dfrac{1}{11}(6+\sqrt{3})\\\\\\d)\ \dfrac{3\sqrt{2}}{\sqrt{2}+2\sqrt{3}}=\dfrac{3\sqrt{2}(\sqrt{2}-2\sqrt{3})}{(\sqrt{2}+2\sqrt{3})(\sqrt{2}-2\sqrt{3})}=\dfrac{3\times2-6\sqrt{6}}{2-12}\\\\\\=\dfrac{6-6\sqrt{6}}{-10}=\dfrac{6(1-\sqrt{6})}{-10}=-\dfrac{3}{5}(1-\sqrt{6})[/tex]
[tex]e)\ \dfrac{2\sqrt{3}}{2\sqrt{3}-5\sqrt{2}}=\dfrac{2\sqrt{3}(2\sqrt{3}+5\sqrt{2})}{(2\sqrt{3}-5\sqrt{2})(2\sqrt{3}+5\sqrt{2})}=\dfrac{4\times3+10\sqrt{6}}{12-50}\\\\\\=\dfrac{12+10\sqrt{6}}{-38}=\dfrac{2(6+5\sqrt{6})}{-38}=-\dfrac{1}{19}(6+5\sqrt{6})[/tex]
1) [tex] \frac{1}{3+ \sqrt{2}} = \frac{1* (3- \sqrt{2}) }{(3+ \sqrt{2})( 3- \sqrt{2})} = \frac{3- \sqrt{2} }{9-2} = \frac{3- \sqrt{2} }{7} [/tex]!
2) [tex] \frac{2}{ \sqrt{3}+ \sqrt{5}} = \frac{2* ( \sqrt{3}+\sqrt{2}) }{( \sqrt{3}- \sqrt{2})( \sqrt{3}+\sqrt{2})= \frac{ 2 (\sqrt{3}+\sqrt{2}) }{3-2} = 2 \sqrt{3}+sqrt{ 2}[/tex]
3)[tex] \frac{ \sqrt{3}}{ 2 \sqrt{3}-1}= \frac{ \sqrt{3} *(2 \sqrt{3}+1) }{ (2 \sqrt{3}-1)(2 \sqrt{3}+1)} = \frac{ 2*3+ /sqrt{3} }{ (4*3)-1} = \frac{ 6+ /sqrt{3} }{ 12-1} = \frac{ 6+ /sqrt{3} }{ 11} [/tex]
4)[tex] \frac{3 \sqrt{2}}{ \sqrt{2}+ 2 \sqrt{3}}= \frac{ 3 \sqrt{2} *( \sqrt{2}- 2 \sqrt{3} ) }{ ( \sqrt{2}+ 2 \sqrt{3} )( \sqrt{2}- 2 \sqrt{3})} = = \frac{ 3*2- 6 \sqrt{6} }{ ( 2-12)} = \frac{ 6(1- \sqrt{6}) }{ -10} \frac{ -3(1- \sqrt{6}) }{ 5} [/tex]
2) [tex] \frac{2}{ \sqrt{3}+ \sqrt{5}} = \frac{2* ( \sqrt{3}+\sqrt{2}) }{( \sqrt{3}- \sqrt{2})( \sqrt{3}+\sqrt{2})= \frac{ 2 (\sqrt{3}+\sqrt{2}) }{3-2} = 2 \sqrt{3}+sqrt{ 2}[/tex]
3)[tex] \frac{ \sqrt{3}}{ 2 \sqrt{3}-1}= \frac{ \sqrt{3} *(2 \sqrt{3}+1) }{ (2 \sqrt{3}-1)(2 \sqrt{3}+1)} = \frac{ 2*3+ /sqrt{3} }{ (4*3)-1} = \frac{ 6+ /sqrt{3} }{ 12-1} = \frac{ 6+ /sqrt{3} }{ 11} [/tex]
4)[tex] \frac{3 \sqrt{2}}{ \sqrt{2}+ 2 \sqrt{3}}= \frac{ 3 \sqrt{2} *( \sqrt{2}- 2 \sqrt{3} ) }{ ( \sqrt{2}+ 2 \sqrt{3} )( \sqrt{2}- 2 \sqrt{3})} = = \frac{ 3*2- 6 \sqrt{6} }{ ( 2-12)} = \frac{ 6(1- \sqrt{6}) }{ -10} \frac{ -3(1- \sqrt{6}) }{ 5} [/tex]