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Pookybear
a fost răspuns

Rationalizati numitorii:
a) 4/√6-2
b) √6+√5/√6-√5
c) 1/3√2-2√5
d) √30-√18/10-2√15


Răspuns :

[tex]\displaystyle a). \frac{4}{ \sqrt{6} -2} = \frac{4( \sqrt{6} +2)}{6-4} = \frac{\not 4 ( \sqrt{6} +2)}{\not2} =2( \sqrt{6} +2)[/tex]

[tex]\displaystyle b) .\frac{ \sqrt{6}+ \sqrt{5} }{ \sqrt{6} - \sqrt{5} } = \frac{( \sqrt{6} + \sqrt{5} )( \sqrt{6} + \sqrt{5} )}{6-5} = \frac{6+ \sqrt{30}+ \sqrt{30}+5 }{1} = \\ =11+2 \sqrt{30}[/tex]

[tex]\displaystyle c). \frac{1}{3 \sqrt{2} -2 \sqrt{5} } = \frac{3 \sqrt{2} +2 \sqrt{5} }{18-20} = \frac{3 \sqrt{2} +2 \sqrt{5} }{-2} =- \frac{3 \sqrt{2} +2 \sqrt{5} }{2} [/tex]

[tex]\displaystyle d). \frac{ \sqrt{30} - \sqrt{18} }{10-2 \sqrt{15} } = \frac{( \sqrt{30} - \sqrt{18} )(10+2 \sqrt{15}) }{100-60} = \\ \\ =\frac{10 \sqrt{30}+2 \sqrt{450}-10 \sqrt{18} -2 \sqrt{270} }{40} = \\ \\ = \frac{10 \sqrt{30}+2 \cdot 15 \sqrt{2} -10 \cdot 3 \sqrt{2} -2 \cdot 3 \sqrt{30} }{40}= \\ \\ = \frac{10 \sqrt{30} +30 \sqrt{2}-30 \sqrt{2} -6 \sqrt{30} }{40} = \frac{\not4 \sqrt{30} }{\not40} = \frac{ \sqrt{30} }{10} [/tex]