Buat soal nomor 2 dari B sampai F...!.... Question from @NoSpam

NamakuDuck B.) $\frac{4-\sqrt{2}}{4+\sqrt{2}}$
=> $(\frac{4-\sqrt{2}}{4+\sqrt{2}})(\frac{4-\sqrt{2}}{4-\sqrt{2}})$
=> $\frac{16-8\sqrt{2}+2}{16-2}$
=> $\frac{18-8\sqrt{2}}{14}$
=> $\frac{9-4\sqrt{2}}{7}$
c.) $\frac{2a}{3a+\sqrt{5}}$
=> $(\frac{2a}{3a+\sqrt{5}})(\frac{3a-\sqrt{5}}{3a-\sqrt{5}})$
=> $\frac{6a^{2}-2a\sqrt{5}}{9a^{2}-5}$
d.) $\frac{\sqrt{3}}{\sqrt{5}-\sqrt{10}}$
=> $(\frac{\sqrt{3}}{\sqrt{5}-\sqrt{10}})(\frac{\sqrt{5}+\sqrt{10}}{\sqrt{5}+\sqrt{10}})$
=> $\frac{\sqrt{15}-\sqrt{30}}{5-10}$
=> $-\frac{\sqrt{15}(1-\sqrt{2})}{5}$
e.) $\frac{xy}{\sqrt{x}+\sqrt{y}}$
=> $(\frac{xy}{\sqrt{x}+\sqrt{y}})(\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}-\sqrt{y}})$
=> $\frac{xy(\sqrt{x}-\sqrt{y})}{x-y}$
f.) $\frac{\sqrt{24}+\sqrt{54}-\sqrt{150}}{\sqrt{96}}$
=> $\frac{2\sqrt{6}+3\sqrt{6}-5\sqrt{6}}{6\sqrt{6}}$
=> $\frac{0}{6\sqrt{6}}$
=> 0

2 votes Thanks 4

handya121 B. $\frac{4- \sqrt{2} }{4+ \sqrt{2} }$
= $\frac{4- \sqrt{2} }{4+ \sqrt{2} } X ( \frac{4- \sqrt{2} }{4- \sqrt{2} } )$
= $\frac{(4- \sqrt{2})(4- \sqrt{2} )}{(4+ \sqrt{2})(4- \sqrt{2} )} }$
= $\frac{16-4 \sqrt{2} -4 \sqrt{2} +2}{16-2}$
= $\frac{18-8 \sqrt{2} }{18}$
= $-8 \sqrt{2}$

C. $\frac{2a}{3a+ \sqrt{5} }$
= $\frac{2a}{3a+ \sqrt{5} } X ( \frac{{3a- \sqrt{5} }}{3a- \sqrt{5} } )$
= $\frac{9a^{2}-2a \sqrt{5} }{9a^{2}+5 }$
= $\frac{-2a \sqrt{5} }{5}$

D. $\frac{ \sqrt{3} }{ \sqrt{5} -\sqrt{10} }$
= $\frac{ \sqrt{3} }{ \sqrt{5} -\sqrt{10} } X( \frac{{ \sqrt{5} +\sqrt{10} }}{{ \sqrt{5} +\sqrt{10} }} )$
= $\frac{ \sqrt{15}+ \sqrt{30} }{5-10}$
= $- \frac{ \sqrt{15} + \sqrt{30} }{5}$

E. $- \frac{xy}{ \sqrt{x}+ \sqrt{y} }$
= $- \frac{xy}{ \sqrt{x}+ \sqrt{y} } X( \frac{ \sqrt{x}- \sqrt{y} }{ \sqrt{x}- \sqrt{y} } )$
= $- \frac{xy \sqrt{x}-xy \sqrt{y} }{x-y}$

F. $\frac{ \sqrt{24} + \sqrt{54} - \sqrt{150} }{ \sqrt{96} }$
= $\frac{ \sqrt{24} + \sqrt{54} - \sqrt{150} }{ \sqrt{96} } X( \frac{ \sqrt{96} }{ \sqrt{96} } )$
= $\frac{2 \sqrt{6}+3 \sqrt{6} -5 \sqrt{6} }{ \sqrt{96} } X( \frac{4 \sqrt{6} }{ \sqrt{96} } )$
= $\frac{48+72-120}{96}$
= $\frac{0}{96}$

1 votes Thanks 3

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