Question

1. Sa se rationalizeze numitorii urmatoarelor fractii:
va rog dau coroana plss​

Answer 1

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

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

$\frac{2}{2\sqrt{3}-1} =\frac{2\left(2\sqrt{3}+1\right)}{\left(2\sqrt{3}-1\right)\left(2\sqrt{3}+1\right)} =\frac{4\sqrt{3}+2}{11}$

$\frac{-5\sqrt{3}+6}{\sqrt{3}-1} =\frac{\left(-5\sqrt{3}+6\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)} =\frac{-9+\sqrt{3}}{2}$

$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} =\frac{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)} =\frac{5+2\sqrt{6}}{1} =5+2\sqrt{6}$

$\frac{\sqrt{3}}{2\sqrt{3}-5\sqrt{2}} =\frac{\sqrt{3}\left(2\sqrt{3}+5\sqrt{2}\right)}{\left(2\sqrt{3}-5\sqrt{2}\right)\left(2\sqrt{3}+5\sqrt{2}\right)} =\frac{6+5\sqrt{6}}{-38} =-\frac{6+5\sqrt{6}}{38}$

Answer 2

Răspuns:

Explicație pas cu pas:

$a) \: \frac{ {}^{ \sqrt{3} + \sqrt{2} )} 2}{ \sqrt{3} - \sqrt{2} } = \frac{2( \sqrt{3} + \sqrt{2} )}{( \sqrt{3} + \sqrt{2})( \sqrt{3} - \sqrt{2}) } = \frac{2 \sqrt{3} + 2 \sqrt{2} }{3 - 2} = 2 \sqrt{3} + 2 \sqrt{2 }$

$b) \: \frac{ {}^{1 + \sqrt{5}) } - 1}{1 - \sqrt{5} } = \frac{ - 1(1 + \sqrt{5}) }{(1 + \sqrt{5})(1 - \sqrt{5} ) } = - \frac{1 + \sqrt{5} }{1 - 5} = - \frac{1 + \sqrt{5} }{ - 4} = \frac{1 +\sqrt{5} }{4}$

$c) \: \frac{ {}^{2 \sqrt{3} + 1)} 3}{2 \sqrt{3} - 1} = \frac{3(2 \sqrt{3} + 1)}{(2 \sqrt{3} + 1)(2 \sqrt{3} - 1) } = \frac{6 \sqrt{3} + 3}{4 \times 3 - 1} = \frac{6 \sqrt{3 } +3 }{11}$

$d) \: \frac{ {}^{ \sqrt{3} + 1) } - 5 \sqrt{3} + 6}{ \sqrt{3} - 1} = \frac{( - 5 \sqrt{3} + 6)( \sqrt{3} + 1) }{( \sqrt{3} + 1)( \sqrt{3} - 1)} = \frac{ - 15 - 5 \sqrt{3} + 6 \sqrt{3} + 6 }{3 - 1} = \frac{ - 9 + \sqrt{3} }{2}$

$e) \: \frac{ {}^{ \sqrt{3} + \sqrt{2}) } \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } = \frac{( \sqrt{3} + \sqrt{2} )( \sqrt{3} + \sqrt{2}) }{( \sqrt{3} - \sqrt{2} )( \sqrt{3} + \sqrt{2} )} = \frac{( \sqrt{3} + \sqrt{2} ) {}^{2} }{3 - 2} = 3 + 2 \sqrt{6} + 2 = 5 + 2 \sqrt{6}$

$f) \: \frac{ {}^{2 \sqrt{3} + 5 \sqrt{2}) } \sqrt{3} }{2 \sqrt{3} - 5 \sqrt{2} } = \frac{ \sqrt{3} (2 \sqrt{3} + 5 \sqrt{2}) }{(2 \sqrt{3} - 5 \sqrt{2})(2 \sqrt{3} + 5 \sqrt{2}) } = \frac{6 + 5 \sqrt{6} }{4 \times 3 - 25 \times 2} = \frac{6 + 5 \sqrt{6} }{12 - 50} = - \frac{ 6 + 5 \sqrt{6} }{38}$