Esercizi sui radicali quadratici doppi

Esercizio

Trasforma in somma o in differenza di due radicali semplici, i seguenti radicali doppi.

1) \sqrt{6 +\sqrt{11}}\sqrt{7 +2\sqrt{6}}

 

2) \sqrt{7 -\sqrt{13}}\sqrt{6 -2\sqrt{5}}

 

3)\sqrt{\frac{2}{3}-\sqrt{\frac{1}{3}}}\sqrt{3a-\sqrt{6a-1}}

 

SVOLGIMENTO

Esercizio

Trasforma in somma o in differenza di due radicali semplici, i seguenti radicali doppi.

1) \sqrt{6 +\sqrt{11}}\sqrt{7 +2\sqrt{6}}

  • \sqrt{6 +\sqrt{11}} applicando la regola \sqrt{a \pm \sqrt{b}} = \sqrt{\frac{a + \sqrt{a^{2}-b}}{2}}  \pm\sqrt{\frac{a - \sqrt{a^{2}-b}}{2}}  abbiamo che:

\sqrt{6 +\sqrt{11}}  = \sqrt{\frac{6+ \sqrt{6^{2}-11}}{2}}  +\sqrt{\frac{6 - \sqrt{6^{2}-11}}{2}} = \sqrt{\frac{6+ \sqrt{36-11}}{2}}  +\sqrt{\frac{6 - \sqrt{36-11}}{2}} =

=\sqrt{\frac{6+ \sqrt{25}}{2}}  +\sqrt{\frac{6 - \sqrt{25}}{2}} =\sqrt{\frac{6+ 5}{2}}  +\sqrt{\frac{6 - 5}{2}}  = \sqrt{\frac{11}{2}}  +\sqrt{\frac{1}{2}}  razionalizziamo = \frac{\sqrt{11}}{\sqrt{2}}+\frac{\sqrt{1}}{\sqrt{2}} =

=\frac{\sqrt{11}\sqrt{2}}{\sqrt{2}\sqrt{2}}+\frac{\sqrt{1}\sqrt{2}}{\sqrt{2}\sqrt{2}} = \frac{\sqrt{22}}{2}+\frac{\sqrt{2}}{2}

  • \sqrt{7 +2\sqrt{6}} = \sqrt{\frac{7+ \sqrt{7^{2}-(2\sqrt{6})^{2}}}{2}}  +\sqrt{\frac{7 - \sqrt{7^{2}-(2\sqrt{6})^{2}}}{2}}   = \sqrt{\frac{7+ \sqrt{49-24}}{2}}  +\sqrt{\frac{7 - \sqrt{49-24}}{2}}  =

=\sqrt{\frac{7+ \sqrt{25}}{2}}  +\sqrt{\frac{7 - \sqrt{25}}{2}}   = \sqrt{\frac{7+ \5}{2}}  +\sqrt{\frac{7 - 5}{2}}   = \sqrt{\frac{12}{2}}  +\sqrt{\frac{2}{2}}    = semplificando otteniamo

\sqrt{6}+1

2) \sqrt{7 -\sqrt{13}}\sqrt{6 -2\sqrt{5}}

  • \sqrt{7 -\sqrt{13}} = \sqrt{\frac{7+ \sqrt{7^{2}-13}}{2}}  -\sqrt{\frac{7 - \sqrt{7^{2}-13}}{2}}  = \sqrt{\frac{7+ \sqrt{49-13}}{2}}  -\sqrt{\frac{7 - \sqrt{49-13}}{2}} =

=\sqrt{\frac{7+ \sqrt{36}}{2}}  -\sqrt{\frac{7 - \sqrt{36}}{2}}  = \sqrt{\frac{7+ 6}{2}}  -\sqrt{\frac{7 - 6}{2}}  = \frac{\sqrt{13}}{\sqrt{2}} – \frac{1}{\sqrt{2}} = razionalizziamo

\frac{\sqrt{13}\sqrt{2}}{\sqrt{2}\sqrt{2}} - \frac{\sqrt{2}}{\sqrt{2}\sqrt{2}}  = \frac{\sqrt{26}}{2} - \frac{\sqrt{2}}{2}  = mettiamo in evidenza 
\frac{\sqrt{2}}{2} (\sqrt{13}-1)

  • \sqrt{6 -2\sqrt{5}}= \sqrt{\frac{6+ \sqrt{6^{2}-(2\sqrt{5})^{2}}}{2}}  -\sqrt{\frac{6 - \sqrt{6^{2}-(2\sqrt{5})^{2}}}{2}}   = \sqrt{\frac{6+ \sqrt{36-20}}{2}}  -\sqrt{\frac{6 - \sqrt{36-20}}{2}}  ==

=\sqrt{\frac{6+ \sqrt{16}}{2}}  -\sqrt{\frac{6 - \sqrt{16}}{2}}  = \sqrt{\frac{6+ 4}{2}}  -\sqrt{\frac{6 - 4}{2}}  = \sqrt{\frac{10}{2}}  -\sqrt{\frac{2}{2}}  = semplificando si ottiene

=\sqrt{5}-1

3)\sqrt{\frac{2}{3}-\sqrt{\frac{1}{3}}}\sqrt{3a-\sqrt{6a-1}}

 

  • \sqrt{\frac{2}{3}-\sqrt{\frac{1}{3}}} = \sqrt{\frac{\frac{2}{3}+ \sqrt{\frac{2}{3}^{2}-\frac{1}{3}}}{2}}  -\sqrt{\frac{\frac{2}{3} - \sqrt{\frac{2}{3}^{2}-\frac{1}{3}}}{2}}  = \sqrt{\frac{\frac{2}{3}+ \sqrt{\frac{4}{9}-\frac{1}{3}}}{2}}  -\sqrt{\frac{\frac{2}{3} - \sqrt{\frac{4}{9}-\frac{1}{3}}}{2}} =

\sqrt{\frac{\frac{2}{3}+ \sqrt{\frac{1}{9}}}{2}}  -\sqrt{\frac{\frac{2}{3} - \sqrt{\frac{1}{9}}}{2}}  = \sqrt{\frac{\frac{2}{3}+ \frac{1}{3}}{2}}  -\sqrt{\frac{\frac{2}{3} -\frac{1}{3}}{2}}  =

=\sqrt{\frac{\frac{3}{3}}{2}}  -\sqrt{\frac{\frac{1}{3} }{2}}  = \sqrt{\frac{1}{2}}  -\sqrt{\frac{1}{6}} }  = razionalizziamo \frac{\sqrt{2}}{\sqrt{2}\sqrt{2}}-\frac{\sqrt{6}}{\sqrt{6}\sqrt{6}} = \frac{\sqrt{2}}{2}-\frac{\sqrt{6}}{6} = \frac{3\sqrt{2}-\sqrt{6}}{6} = \frac{\sqrt{2}}{6}(3-\sqrt{3})

  • \sqrt{3a-\sqrt{6a-1}} = \sqrt{\frac{3a+ \sqrt{9a^{2}-(6a-1)}}{2}}  -\sqrt{\frac{3a - \sqrt{9a^{2}-(6a.
-1)}}{2}}   =

\sqrt{\frac{3a+ \sqrt{9a^{2}-6a+1}}{2}}  -\sqrt{\frac{3a - \sqrt{9a^{2}-6a+1}}{2}}   = \sqrt{\frac{3a+ \sqrt{(3a-1)^{2}}}{2}}  -\sqrt{\frac{3a - \sqrt{(3a-1)^{2}}}{2}}  =

\sqrt{\frac{3a+3a-1}{2}}  -\sqrt{\frac{3a - (3a-1)}{2}}   = \sqrt{\frac{6a-1}{2}}  -\sqrt{\frac{3a - 3a+1}{2}}   = \sqrt{\frac{6a-1}{2}}  -\sqrt{\frac{+1}{2}}   =

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