Racine carrée/Exercices/Quantité conjuguée

Exercice 2-1

Simplifier les expressions suivantes :

Solution

$\left(5+2{\sqrt {7}}\right)\left(5-2{\sqrt {7}}\right)=5^{2}-\left(2{\sqrt {7}}\right)^{2}=25-4\times 7=-3$ ;
${\sqrt {2+{\sqrt {3}}}}\,{\sqrt {2-{\sqrt {3}}}}={\sqrt {\left(2+{\sqrt {3}}\right)\left(2-{\sqrt {3}}\right)}}={\sqrt {2^{2}-{\sqrt {3}}^{2}}}={\sqrt {1}}=1$ .

Rendre rationnels les dénominateurs des fractions suivantes :

Solution

${\frac {1}{\sqrt {8}}}={\frac {\sqrt {2}}{{\sqrt {2}}{\sqrt {8}}}}={\frac {\sqrt {2}}{\sqrt {16}}}={\frac {\sqrt {2}}{4}}$ ;
${\frac {1}{{\sqrt {2}}-1}}={\frac {{\sqrt {2}}+1}{\left({\sqrt {2}}+1\right)\left({\sqrt {2}}-1\right)}}={\frac {{\sqrt {2}}+1}{{\sqrt {2}}^{2}-1^{2}}}={\sqrt {2}}+1$ ;
${\frac {{\sqrt {3}}-{\sqrt {2}}}{{\sqrt {3}}+{\sqrt {2}}}}={\frac {\left({\sqrt {3}}-{\sqrt {2}}\right)^{2}}{\left({\sqrt {3}}+{\sqrt {2}}\right)\left({\sqrt {3}}-{\sqrt {2}}\right)}}={\frac {{\sqrt {3}}^{2}-2{\sqrt {3}}{\sqrt {2}}+{\sqrt {2}}^{2}}{{\sqrt {3}}^{2}-{\sqrt {2}}^{2}}}={\frac {3-2{\sqrt {6}}+2}{3-2}}=5-2{\sqrt {6}}$ ;
${\frac {3+{\sqrt {5}}}{{\sqrt {5}}-1}}={\frac {\left(3+{\sqrt {5}}\right)\left({\sqrt {5}}+1\right)}{\left({\sqrt {5}}-1\right)\left({\sqrt {5}}+1\right)}}={\frac {3{\sqrt {5}}+3+{\sqrt {5}}^{2}+{\sqrt {5}}}{{\sqrt {5}}^{2}-1^{2}}}={\frac {8+4{\sqrt {5}}}{4}}=2+{\sqrt {5}}$ .

Effectuer les sommes suivantes :

${\frac {{\sqrt {2}}+{\sqrt {3}}}{{\sqrt {3}}-{\sqrt {2}}}}+{\frac {{\sqrt {2}}-{\sqrt {3}}}{{\sqrt {2}}+{\sqrt {3}}}}$ ;
${\frac {1}{{\sqrt {3}}+6}}+{\frac {1}{{\sqrt {3}}-6}}$ .

Solution

${\frac {{\sqrt {2}}+{\sqrt {3}}}{{\sqrt {3}}-{\sqrt {2}}}}+{\frac {{\sqrt {2}}-{\sqrt {3}}}{{\sqrt {2}}+{\sqrt {3}}}}={\frac {\left({\sqrt {3}}+{\sqrt {2}}\right)^{2}-\left({\sqrt {3}}-{\sqrt {2}}\right)^{2}}{\left({\sqrt {3}}+{\sqrt {2}}\right)\left({\sqrt {3}}-{\sqrt {2}}\right)}}={\frac {\left({\sqrt {3}}+{\sqrt {2}}+{\sqrt {3}}-{\sqrt {2}}\right)\left({\sqrt {3}}+{\sqrt {2}}-\left({\sqrt {3}}-{\sqrt {2}}\right)\right)}{{\sqrt {3}}^{2}-{\sqrt {2}}^{2}}}={\frac {2{\sqrt {3}}\;2{\sqrt {2}}}{3-2}}=4{\sqrt {6}}$ ;
${\frac {1}{{\sqrt {3}}+6}}+{\frac {1}{{\sqrt {3}}-6}}={\frac {{\sqrt {3}}-6+{\sqrt {3}}+6}{\left({\sqrt {3}}+6\right)\left({\sqrt {3}}-6\right)}}={\frac {2{\sqrt {3}}}{{\sqrt {3}}^{2}-6^{2}}}=-{\frac {2{\sqrt {3}}}{-33}}$ .

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