Expressions algébriques/Exercices/Expressions irrationnelles

Dans toute la page, même si ce n'est pas indiqué, les variables prennent des valeurs telles que les calculs soient définis.

Exercice 7-1

Rendre rationnel le dénominateur des expressions suivantes :

a)  ${\frac {1}{2{\sqrt {2}}+{\sqrt {3}}+{\sqrt {10}}+{\sqrt {15}}}}$

b)  ${\frac {1}{1+{\sqrt[{3}]{2}}+{\sqrt[{3}]{4}}}}$

c)  ${\frac {1}{{\sqrt {2}}-{\sqrt {3}}+{\sqrt {5}}-{\sqrt {7}}}}$

d)  ${\frac {1}{{\sqrt[{3}]{2}}+{\sqrt[{3}]{3}}+{\sqrt[{3}]{5}}}}$

a) Corrigé.

${\begin{aligned}{\frac {1}{2{\sqrt {2}}+{\sqrt {3}}+{\sqrt {10}}+{\sqrt {15}}}}&={\frac {(2{\sqrt {2}}+{\sqrt {10}})-({\sqrt {3}}+{\sqrt {15}})}{\left[(2{\sqrt {2}}+{\sqrt {10}})+({\sqrt {3}}+{\sqrt {15}})\right]\left[(2{\sqrt {2}}+{\sqrt {10}})-({\sqrt {3}}+{\sqrt {15}})\right]}}\\&={\frac {2{\sqrt {2}}+{\sqrt {10}}-{\sqrt {3}}-{\sqrt {15}}}{(2{\sqrt {2}}+{\sqrt {10}})^{2}-({\sqrt {3}}+{\sqrt {15}})^{2}}}\\&={\frac {2{\sqrt {2}}+{\sqrt {10}}-{\sqrt {3}}-{\sqrt {15}}}{(8+8{\sqrt {5}}+10)-(3+6{\sqrt {5}}+15)}}\\&={\frac {2{\sqrt {2}}+{\sqrt {10}}-{\sqrt {3}}-{\sqrt {15}}}{2{\sqrt {5}}}}\\&={\frac {(2{\sqrt {2}}+{\sqrt {10}}-{\sqrt {3}}-{\sqrt {15}}){\sqrt {5}}}{2{\sqrt {5}}{\sqrt {5}}}}\\&={\frac {2{\sqrt {10}}+5{\sqrt {2}}-{\sqrt {15}}-5{\sqrt {3}}}{10}}\\\end{aligned}}$

 

b) Corrigé.

Dans l'identité remarquable :

$(a+b+c)(a^{2}+b^{2}+c^{2}-ab-ac-bc)=a^{3}+b^{3}+c^{3}-3abc$

nous poserons

$a=1\qquad \qquad b={\sqrt[{3}]{2}}\qquad \qquad c={\sqrt[{3}]{4}}$

et multiplions le numérateur et le dénominateur par $a^{2}+b^{2}+c^{2}-ab-ac-bc$. On obtient :

${\begin{aligned}{\frac {1}{1+{\sqrt[{3}]{2}}+{\sqrt[{3}]{4}}}}&={\frac {a^{2}+b^{2}+c^{2}-ab-ac-bc}{\left(1+{\sqrt[{3}]{2}}+{\sqrt[{3}]{4}}\right)(a^{2}+b^{2}+c^{2}-ab-ac-bc)}}\\&={\frac {1+{\sqrt[{3}]{4}}+{\sqrt[{3}]{16}}-{\sqrt[{3}]{2}}-{\sqrt[{3}]{4}}-{\sqrt[{3}]{8}}}{1+2+4-3{\sqrt[{3}]{1\times 2\times 4}}}}\\&={\frac {1+{\sqrt[{3}]{4}}+2{\sqrt[{3}]{2}}-{\sqrt[{3}]{2}}-{\sqrt[{3}]{4}}-2}{7-3{\sqrt[{3}]{8}}}}\\&={\frac {{\sqrt[{3}]{2}}-1}{7-3\times 2}}\\&={\sqrt[{3}]{2}}-1\end{aligned}}$

 

c) Corrigé.

${\begin{aligned}{\frac {1}{{\sqrt {2}}-{\sqrt {3}}+{\sqrt {5}}-{\sqrt {7}}}}&={\frac {({\sqrt {2}}-{\sqrt {3}})-({\sqrt {5}}-{\sqrt {7}})}{\left[({\sqrt {2}}-{\sqrt {3}})+({\sqrt {5}}-{\sqrt {7}})\right]\left[({\sqrt {2}}-{\sqrt {3}})-({\sqrt {5}}-{\sqrt {7}})\right]}}\\&={\frac {({\sqrt {2}}-{\sqrt {3}})-({\sqrt {5}}-{\sqrt {7}})}{({\sqrt {2}}-{\sqrt {3}})^{2}-({\sqrt {5}}-{\sqrt {7}})^{2}}}\\&={\frac {{\sqrt {2}}-{\sqrt {3}}-{\sqrt {5}}+{\sqrt {7}}}{(5-2{\sqrt {6}})-(12-2{\sqrt {35}})}}\\&={\frac {{\sqrt {2}}-{\sqrt {3}}-{\sqrt {5}}+{\sqrt {7}}}{2{\sqrt {35}}-7-2{\sqrt {6}}}}\\&={\frac {\left[{\sqrt {2}}-{\sqrt {3}}-{\sqrt {5}}+{\sqrt {7}}\right]\left[2{\sqrt {35}}+7+2{\sqrt {6}}\right]}{\left[2{\sqrt {35}}-(7+2{\sqrt {6}})\right]\left[2{\sqrt {35}}+(7+2{\sqrt {6}})\right]}}\\&={\frac {2{\sqrt {70}}+7{\sqrt {2}}+4{\sqrt {3}}-2{\sqrt {105}}-7{\sqrt {3}}-6{\sqrt {2}}-10{\sqrt {7}}-7{\sqrt {5}}-2{\sqrt {30}}+14{\sqrt {5}}+7{\sqrt {7}}+2{\sqrt {42}}}{140-(7+2{\sqrt {6}})^{2}}}\\&={\frac {{\sqrt {2}}-3{\sqrt {3}}+7{\sqrt {5}}-3{\sqrt {7}}-2{\sqrt {30}}+2{\sqrt {42}}+2{\sqrt {70}}-2{\sqrt {105}}}{67-28{\sqrt {6}}}}\\&={\frac {\left({\sqrt {2}}-3{\sqrt {3}}+7{\sqrt {5}}-3{\sqrt {7}}-2{\sqrt {30}}+2{\sqrt {42}}+2{\sqrt {70}}-2{\sqrt {105}}\right)\left(67+28{\sqrt {6}}\right)}{\left(67-28{\sqrt {6}}\right)\left(67+28{\sqrt {6}}\right)}}\\&={\frac {-185{\sqrt {2}}-145{\sqrt {3}}+133{\sqrt {5}}+135{\sqrt {7}}+62{\sqrt {30}}+50{\sqrt {42}}-34{\sqrt {70}}-22{\sqrt {105}}}{-215}}\\&={\frac {185{\sqrt {2}}+145{\sqrt {3}}-133{\sqrt {5}}-135{\sqrt {7}}-62{\sqrt {30}}-50{\sqrt {42}}+34{\sqrt {70}}+22{\sqrt {105}}}{215}}\\\end{aligned}}$

 

d) Corrigé.

Dans l'identité remarquable :

$(a+b+c)(a^{2}+b^{2}+c^{2}-ab-ac-bc)=a^{3}+b^{3}+c^{3}-3abc$

nous poserons

$a={\sqrt[{3}]{2}}\qquad \qquad b={\sqrt[{3}]{3}}\qquad \qquad c={\sqrt[{3}]{5}}$

et multiplions le numérateur et le dénominateur par $a^{2}+b^{2}+c^{2}-ab-ac-bc$. On obtient :

${\begin{aligned}{\frac {1}{{\sqrt[{3}]{2}}+{\sqrt[{3}]{3}}+{\sqrt[{3}]{5}}}}&={\frac {a^{2}+b^{2}+c^{2}-ab-ac-bc}{\left({\sqrt[{3}]{2}}+{\sqrt[{3}]{3}}+{\sqrt[{3}]{5}}\right)(a^{2}+b^{2}+c^{2}-ab-ac-bc)}}\\&={\frac {{\sqrt[{3}]{4}}+{\sqrt[{3}]{9}}+{\sqrt[{3}]{25}}-{\sqrt[{3}]{6}}-{\sqrt[{3}]{10}}-{\sqrt[{3}]{15}}}{2+3+5-3{\sqrt[{3}]{2\times 3\times 5}}}}\\&={\frac {{\sqrt[{3}]{4}}+{\sqrt[{3}]{9}}+{\sqrt[{3}]{25}}-{\sqrt[{3}]{6}}-{\sqrt[{3}]{10}}-{\sqrt[{3}]{15}}}{10-3{\sqrt[{3}]{30}}}}\end{aligned}}$

Nous allons maintenant utiliser l'identité remarquable :

$a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$

en posant :

$a=10\qquad \qquad b=3{\sqrt[{3}]{30}}$

Nous continuerons ainsi :

${\begin{aligned}\qquad &={\frac {\left({\sqrt[{3}]{4}}+{\sqrt[{3}]{9}}+{\sqrt[{3}]{25}}-{\sqrt[{3}]{6}}-{\sqrt[{3}]{10}}-{\sqrt[{3}]{15}}\right)(a^{2}+ab+b^{2})}{\left(10-3{\sqrt[{3}]{30}}\right)(a^{2}+ab+b^{2})}}\\&={\frac {\left({\sqrt[{3}]{4}}+{\sqrt[{3}]{9}}+{\sqrt[{3}]{25}}-{\sqrt[{3}]{6}}-{\sqrt[{3}]{10}}-{\sqrt[{3}]{15}}\right)\left(100+30{\sqrt[{3}]{3}}0+9{\sqrt[{3}]{900}}\right))}{10^{3}-\left(3{\sqrt[{3}]{30}}\right)^{3}}}\\&={\frac {100{\sqrt[{3}]{4}}+60{\sqrt[{3}]{15}}+18{\sqrt[{3}]{450}}+100{\sqrt[{3}]{9}}+90{\sqrt[{3}]{10}}+27{\sqrt[{3}]{300}}+100{\sqrt[{3}]{25}}+150{\sqrt[{3}]{6}}+45{\sqrt[{3}]{180}}-100{\sqrt[{3}]{6}}-30{\sqrt[{3}]{180}}-54{\sqrt[{3}]{25}}-100{\sqrt[{3}]{10}}-30{\sqrt[{3}]{300}}-90{\sqrt[{3}]{9}}-100{\sqrt[{3}]{15}}-30{\sqrt[{3}]{450}}-135{\sqrt[{3}]{4}}}{1000-810}}\\&={\frac {-35{\sqrt[{3}]{4}}+50{\sqrt[{3}]{6}}+10{\sqrt[{3}]{9}}-10{\sqrt[{3}]{10}}-40{\sqrt[{3}]{15}}+46{\sqrt[{3}]{25}}+15{\sqrt[{3}]{180}}-3{\sqrt[{3}]{300}}-12{\sqrt[{3}]{450}}}{190}}\end{aligned}}$

 

Exercice 7-2

Établir les relations suivantes :

a)  $\left({\frac {11}{5-{\sqrt {3}}}}\right)^{2}-\left({\frac {5-2{\sqrt {5}}}{2-{\sqrt {5}}}}\right)^{2}={\sqrt {{\frac {91}{4}}+10{\sqrt {3}}}}$

b)  ${\frac {2{\sqrt {9+{\sqrt {65}}}}}{{\sqrt {19}}-{\sqrt {3}}}}={\frac {{\sqrt {19}}+{\sqrt {3}}}{2{\sqrt {9-{\sqrt {65}}}}}}$

c)  ${\sqrt {8+2{\sqrt {10+2{\sqrt {5}}}}}}+{\sqrt {8-2{\sqrt {10+2{\sqrt {5}}}}}}={\sqrt {2}}\left({\sqrt {5}}+1\right)$

d)  ${\frac {{\sqrt {3}}-1}{{\sqrt {3}}+1}}={\sqrt[{3}]{\frac {9-5{\sqrt {3}}}{9+5{\sqrt {3}}}}}$

e)  ${\frac {1+{\sqrt {3}}}{2{\sqrt[{3}]{2}}}}={\frac {2+{\sqrt {3}}}{\sqrt[{3}]{20+12{\sqrt {3}}}}}$

f)  ${\sqrt[{3}]{38+17{\sqrt {5}}}}={\sqrt {9+4{\sqrt {5}}}}$

a) Corrigé.

${\begin{aligned}\left({\frac {11}{5-{\sqrt {3}}}}\right)^{2}-\left({\frac {5-2{\sqrt {5}}}{2-{\sqrt {5}}}}\right)^{2}&=\left({\frac {11(5+{\sqrt {3}})}{(5-{\sqrt {3}})(5+{\sqrt {3}})}}\right)^{2}-\left({\frac {(5-2{\sqrt {5}})(2+{\sqrt {5}})}{(2-{\sqrt {5}})(2+{\sqrt {5}})}}\right)^{2}\\&=\left({\frac {11(5+{\sqrt {3}})}{22}}\right)^{2}-\left({\frac {10+5{\sqrt {5}}-4{\sqrt {5}}-2\times 5}{-1}}\right)^{2}\\&=\left({\frac {5+{\sqrt {3}}}{2}}\right)^{2}-\left(-{\sqrt {5}}\right)^{2}\\&={\frac {25+10{\sqrt {3}}+3}{4}}-{\frac {20}{4}}\\&={\frac {8+10{\sqrt {3}}}{4}}\\&={\frac {4+5{\sqrt {3}}}{2}}\\\end{aligned}}$

On est donc ramené à vérifier que :

${\frac {4+5{\sqrt {3}}}{2}}={\sqrt {{\frac {91}{4}}+10{\sqrt {3}}}}$

Ce qui se fait aisément en remarquant que :

$\left({\frac {4+5{\sqrt {3}}}{2}}\right)^{2}={\frac {16+2\times 4\times 5{\sqrt {3}}+75}{4}}={\frac {91+40{\sqrt {3}}}{4}}={\frac {91}{4}}+10{\sqrt {3}}$

 

b) Corrigé.

La relation est équivalente à :

$4{\sqrt {(9+{\sqrt {65}})(9-{\sqrt {65}})}}=({\sqrt {19}}+{\sqrt {3}})({\sqrt {19}}-{\sqrt {3}})$

qui s'écrit :

$4{\sqrt {81-65}}=19-3$

soit

${\sqrt {16}}=4$

Ce qui est vrai.

 

c) Corrigé.

Comme les deux membres sont positifs, montrons que le carré des deux membres sont égaux.

Le carré du premier membre est :

${\begin{aligned}\left({\sqrt {8+2{\sqrt {10+2{\sqrt {5}}}}}}+{\sqrt {8-2{\sqrt {10+2{\sqrt {5}}}}}}\right)^{2}&=8+2{\sqrt {10+2{\sqrt {5}}}}+2{\sqrt {8+2{\sqrt {10+2{\sqrt {5}}}}}}{\sqrt {8-2{\sqrt {10+2{\sqrt {5}}}}}}+8-2{\sqrt {10+2{\sqrt {5}}}}\\&=16+2{\sqrt {64-4(10+2{\sqrt {5}})}}\\&=16+2{\sqrt {24-8{\sqrt {5}}}}\\&=16+4{\sqrt {6-2{\sqrt {5}}}}\\\end{aligned}}$

Le carré du deuxième membre est :

$\left({\sqrt {2}}\left({\sqrt {5}}+1\right)\right)^{2}=2\left(6+2{\sqrt {5}}\right)$

On s'est donc ramené à établir la relation plus simple :

$16+4{\sqrt {6-2{\sqrt {5}}}}=2\left(6+2{\sqrt {5}}\right)$

qui s'écrit encore plus simplement :

${\sqrt {6-2{\sqrt {5}}}}={\sqrt {5}}-1$

Qui est vraie puisque :

$\left({\sqrt {5}}-1\right)^{2}=6-2{\sqrt {5}}$

 

d) Corrigé.

Il nous suffit de montrer que :

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

Effectivement, on a bien :

${\begin{aligned}\left({\frac {{\sqrt {3}}-1}{{\sqrt {3}}+1}}\right)^{3}&={\frac {({\sqrt {3}})^{3}-3({\sqrt {3}})^{2}+3{\sqrt {3}}-1^{3}}{({\sqrt {3}})^{3}+3({\sqrt {3}})^{2}+3{\sqrt {3}}+1^{3}}}\\&={\frac {3{\sqrt {3}}-9+3{\sqrt {3}}-1}{3{\sqrt {3}}+9+3{\sqrt {3}}+1}}\\&={\frac {6{\sqrt {3}}-10}{6{\sqrt {3}}+10}}\\&={\frac {3{\sqrt {3}}-5}{3{\sqrt {3}}+5}}\\&={\frac {{\sqrt {3}}\left(3{\sqrt {3}}-5\right)}{{\sqrt {3}}\left(3{\sqrt {3}}+5\right)}}\\&={\frac {9-5{\sqrt {3}}}{9+5{\sqrt {3}}}}\end{aligned}}$

 

e) Corrigé.

Calculons le cube des deux membres :

Pour le premier membre :

${\begin{aligned}\left({\frac {1+{\sqrt {3}}}{2{\sqrt[{3}]{2}}}}\right)^{3}&={\frac {1^{3}+3{\sqrt {3}}+3({\sqrt {3}})^{2}+({\sqrt {3}})^{3}}{(2{\sqrt[{3}]{2}})^{3}}}\\&={\frac {1+3{\sqrt {3}}+9+3{\sqrt {3}}}{16}}\\&={\frac {10+6{\sqrt {3}}}{16}}\\&={\frac {5+3{\sqrt {3}}}{8}}\\\end{aligned}}$

Pour le deuxième membre :

${\begin{aligned}\left({\frac {2+{\sqrt {3}}}{\sqrt[{3}]{20+12{\sqrt {3}}}}}\right)^{3}&={\frac {2^{3}+3\times 2^{2}{\sqrt {3}}+3\times 2({\sqrt {3}})^{2}+({\sqrt {3}})^{3}}{20+12{\sqrt {3}}}}\\&={\frac {8+12{\sqrt {3}}+18+3{\sqrt {3}}}{20+12{\sqrt {3}}}}\\&={\frac {26+15{\sqrt {3}}}{20+12{\sqrt {3}}}}\\&={\frac {(26+15{\sqrt {3}})(20-12{\sqrt {3}})}{(20+12{\sqrt {3}})(20-12{\sqrt {3}})}}\\&={\frac {520-312{\sqrt {3}}+300{\sqrt {3}}-540}{400-432}}\\&={\frac {-20-12{\sqrt {3}}}{-32}}\\&={\frac {5+3{\sqrt {3}}}{8}}\end{aligned}}$

En élevant les deux membres au cube, nous obtenons le même résultat. On a donc bien :

${\frac {1+{\sqrt {3}}}{2{\sqrt[{3}]{2}}}}={\frac {2+{\sqrt {3}}}{\sqrt[{3}]{20+12{\sqrt {3}}}}}$

 

f) Corrigé.

Pour établir cette égalité, nous comparerons les deux membres élevées à la puissance 6.

Pour le premier membre, on a :

${\begin{aligned}\left({\sqrt[{3}]{38+17{\sqrt {5}}}}\right)^{6}&=\left(38+17{\sqrt {5}}\right)^{2}\\&=38^{2}+2\times 38\times 17{\sqrt {5}}+\left(17{\sqrt {5}}\right)^{2}\\&=2889+1292{\sqrt {5}}\end{aligned}}$

Pour le deuxième membre, on a :

${\begin{aligned}\left({\sqrt {9+4{\sqrt {5}}}}\right)^{6}&=\left(9+4{\sqrt {5}}\right)^{3}\\&=9^{3}+3\times 9^{2}\times 4{\sqrt {5}}+3\times 9\times \left(4{\sqrt {5}}\right)^{2}+\left(4{\sqrt {5}}\right)^{3}\\&=2889+1292{\sqrt {5}}\end{aligned}}$

Les deux membres à la puissance 6 étant égaux, on a bien :

${\sqrt[{3}]{38+17{\sqrt {5}}}}={\sqrt {9+4{\sqrt {5}}}}$

 

Exercice 7-3

Simplifier les expressions suivantes :

a)  ${\sqrt {\frac {5+2{\sqrt {6}}}{3+2{\sqrt {2}}}}}$

b)  ${\sqrt {\frac {7+2{\sqrt {10}}}{5+2{\sqrt {6}}}}}$

c)  ${\sqrt {\frac {5-{\sqrt {6}}}{13+5{\sqrt {6}}}}}$

d)  ${\sqrt {\frac {4+{\sqrt {7}}}{3+{\sqrt {5}}}}}$

e)  ${\frac {2+{\sqrt {3}}}{{\sqrt {2}}+{\sqrt {2+{\sqrt {3}}}}}}+{\frac {2-{\sqrt {3}}}{{\sqrt {2}}-{\sqrt {2-{\sqrt {3}}}}}}$

f)  ${\frac {\sqrt {3-2{\sqrt {2}}}}{\sqrt {17-12{\sqrt {2}}}}}-{\frac {\sqrt {3+2{\sqrt {2}}}}{\sqrt {17+12{\sqrt {2}}}}}$

h)  ${\frac {{\sqrt {{\sqrt[{4}]{8}}+{\sqrt {{\sqrt {2}}-1}}}}-{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}-1}}}}}{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}+1}}}}}$

a) Corrigé.

${\begin{aligned}{\sqrt {\frac {5+2{\sqrt {6}}}{3+2{\sqrt {2}}}}}&={\sqrt {\frac {3+2{\sqrt {2}}{\sqrt {3}}+2}{2+2{\sqrt {2}}+1}}}\\&={\sqrt {\frac {\left({\sqrt {3}}\right)^{2}+2{\sqrt {2}}{\sqrt {3}}+\left({\sqrt {2}}\right)^{2}}{\left({\sqrt {2}}\right)^{2}+2{\sqrt {2}}+1}}}\\&={\sqrt {\frac {\left({\sqrt {3}}+{\sqrt {2}}\right)^{2}}{\left({\sqrt {2}}+1\right)^{2}}}}\\&={\frac {{\sqrt {3}}+{\sqrt {2}}}{{\sqrt {2}}+1}}\\&={\frac {({\sqrt {3}}+{\sqrt {2}})({\sqrt {2}}-1)}{({\sqrt {2}}+1)({\sqrt {2}}-1)}}\\&=({\sqrt {3}}+{\sqrt {2}})({\sqrt {2}}-1)\end{aligned}}$

 

b) Corrigé.

${\begin{aligned}{\sqrt {\frac {7+2{\sqrt {10}}}{5+2{\sqrt {6}}}}}&={\sqrt {\frac {5+2{\sqrt {2}}{\sqrt {5}}+2}{3+2{\sqrt {2}}{\sqrt {3}}+2}}}\\&={\sqrt {\frac {\left({\sqrt {5}}\right)^{2}+2{\sqrt {2}}{\sqrt {5}}+\left({\sqrt {2}}\right)^{2}}{\left({\sqrt {3}}\right)^{2}+2{\sqrt {2}}{\sqrt {3}}+\left({\sqrt {2}}\right)^{2}}}}\\&={\sqrt {\frac {\left({\sqrt {5}}+{\sqrt {2}}\right)^{2}}{\left({\sqrt {3}}+{\sqrt {2}}\right)^{2}}}}\\&={\frac {{\sqrt {5}}+{\sqrt {2}}}{{\sqrt {3}}+{\sqrt {2}}}}\\&={\frac {({\sqrt {5}}+{\sqrt {2}})({\sqrt {3}}-{\sqrt {2}})}{({\sqrt {3}}+{\sqrt {2}})({\sqrt {3}}-{\sqrt {2}})}}\\&={\frac {({\sqrt {5}}+{\sqrt {2}})({\sqrt {3}}-{\sqrt {2}})}{3-2}}\\&=({\sqrt {5}}+{\sqrt {2}})({\sqrt {3}}-{\sqrt {2}})\end{aligned}}$

 

c) Corrigé.

${\begin{aligned}{\sqrt {\frac {5-{\sqrt {6}}}{13+5{\sqrt {6}}}}}&={\sqrt {\frac {(5-{\sqrt {6}})(13-5{\sqrt {6}})}{(13+5{\sqrt {6}})(13-5{\sqrt {6}})}}}\\&={\sqrt {\frac {65-25{\sqrt {6}}-13{\sqrt {6}}+30}{169-150}}}\\&={\sqrt {\frac {95-38{\sqrt {6}}}{19}}}\\&={\sqrt {5-2{\sqrt {6}}}}\\&={\sqrt {2-2{\sqrt {6}}+3}}\\&={\sqrt {\left({\sqrt {2}}+{\sqrt {3}}\right)^{2}}}\\&={\sqrt {2}}+{\sqrt {3}}\end{aligned}}$

 

d) Corrigé.

${\begin{aligned}{\sqrt {\frac {4+{\sqrt {7}}}{3+{\sqrt {5}}}}}&={\sqrt {\frac {8+2{\sqrt {7}}}{6+2{\sqrt {5}}}}}\\&={\sqrt {\frac {7+2{\sqrt {7}}+1}{5+2{\sqrt {5}}+1}}}\\&={\sqrt {\frac {\left({\sqrt {7}}+1\right)^{2}}{\left({\sqrt {5}}+1\right)^{2}}}}\\&={\frac {{\sqrt {7}}+1}{{\sqrt {5}}+1}}\\&={\frac {({\sqrt {7}}+1)({\sqrt {5}}-1)}{({\sqrt {5}}+1)({\sqrt {5}}-1)}}\\&={\frac {{\sqrt {35}}-{\sqrt {7}}+{\sqrt {5}}-1}{4}}\end{aligned}}$

 

e) Corrigé.

${\begin{aligned}{\frac {2+{\sqrt {3}}}{{\sqrt {2}}+{\sqrt {2+{\sqrt {3}}}}}}+{\frac {2-{\sqrt {3}}}{{\sqrt {2}}-{\sqrt {2-{\sqrt {3}}}}}}&={\frac {2+{\sqrt {3}}}{{\sqrt {2}}+{\sqrt {\frac {3+2{\sqrt {3}}+1}{2}}}}}+{\frac {2-{\sqrt {3}}}{{\sqrt {2}}-{\sqrt {\frac {3-2{\sqrt {3}}+1}{2}}}}}\\&={\frac {2+{\sqrt {3}}}{{\sqrt {2}}+{\frac {\sqrt {\left({\sqrt {3}}+1\right)^{2}}}{\sqrt {2}}}}}+{\frac {2-{\sqrt {3}}}{{\sqrt {2}}-{\frac {\sqrt {\left({\sqrt {3}}-1\right)^{2}}}{\sqrt {2}}}}}\\&={\frac {2+{\sqrt {3}}}{{\sqrt {2}}+{\frac {{\sqrt {3}}+1}{\sqrt {2}}}}}+{\frac {2-{\sqrt {3}}}{{\sqrt {2}}-{\frac {{\sqrt {3}}-1}{\sqrt {2}}}}}\\&={\frac {2{\sqrt {2}}+{\sqrt {6}}}{2+{\sqrt {3}}+1}}+{\frac {2{\sqrt {2}}-{\sqrt {6}}}{2-{\sqrt {3}}+1}}\\&={\frac {2{\sqrt {2}}+{\sqrt {6}}}{3+{\sqrt {3}}}}+{\frac {2{\sqrt {2}}-{\sqrt {6}}}{3-{\sqrt {3}}}}\\&={\frac {(2{\sqrt {2}}+{\sqrt {6}})(3-{\sqrt {3}})+(2{\sqrt {2}}-{\sqrt {6}})(3+{\sqrt {3}})}{(3+{\sqrt {3}})(3-{\sqrt {3}})}}\\&={\frac {6{\sqrt {2}}-2{\sqrt {6}}+3{\sqrt {6}}-3{\sqrt {2}}+6{\sqrt {2}}+2{\sqrt {6}}-3{\sqrt {6}}-3{\sqrt {2}}}{9-3}}\\&={\frac {6{\sqrt {2}}}{6}}\\&={\sqrt {2}}\end{aligned}}$

 

f) Corrigé.

${\begin{aligned}{\frac {\sqrt {3-2{\sqrt {2}}}}{\sqrt {17-12{\sqrt {2}}}}}-{\frac {\sqrt {3+2{\sqrt {2}}}}{\sqrt {17+12{\sqrt {2}}}}}&={\frac {\sqrt {2-2{\sqrt {2}}+1}}{\sqrt {9-2\times 3{\sqrt {8}}+8}}}-{\frac {\sqrt {2+2{\sqrt {2}}+1}}{\sqrt {9+2\times 3{\sqrt {8}}+8}}}\\&={\frac {\sqrt {\left({\sqrt {2}}-1\right)^{2}}}{\sqrt {\left(3-{\sqrt {8}}\right)^{2}}}}-{\frac {\sqrt {\left({\sqrt {2}}+1\right)^{2}}}{\sqrt {\left(3+{\sqrt {8}}\right)^{2}}}}\\&={\frac {{\sqrt {2}}-1}{3-{\sqrt {8}}}}-{\frac {{\sqrt {2}}+1}{3+{\sqrt {8}}}}\\&={\frac {({\sqrt {2}}-1)(3+{\sqrt {8}})-({\sqrt {2}}+1)(3-{\sqrt {8}})}{(3-{\sqrt {8}})(3+{\sqrt {8}})}}\\&={\frac {3{\sqrt {2}}+4-3-{\sqrt {8}}-3{\sqrt {2}}+4-3+{\sqrt {8}}}{9-8}}\\&=2\end{aligned}}$

 

h) Corrigé.

${\begin{aligned}{\frac {{\sqrt {{\sqrt[{4}]{8}}+{\sqrt {{\sqrt {2}}-1}}}}-{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}-1}}}}}{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}+1}}}}}&={\frac {\sqrt {\left({\sqrt {{\sqrt[{4}]{8}}+{\sqrt {{\sqrt {2}}-1}}}}-{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}-1}}}}\right)^{2}}}{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}+1}}}}}\\&={\frac {\sqrt {{\sqrt[{4}]{8}}+{\sqrt {{\sqrt {2}}-1}}-2{\sqrt {{\sqrt[{4}]{8}}+{\sqrt {{\sqrt {2}}-1}}}}{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}-1}}}}+{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}-1}}}}{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}+1}}}}}\\&={\frac {\sqrt {2{\sqrt[{4}]{8}}-2{\sqrt {\left({\sqrt[{4}]{8}}+{\sqrt {{\sqrt {2}}-1}}\right)\left({\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}-1}}\right)}}}}{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}+1}}}}}\\&={\sqrt {2}}{\frac {\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {8}}-\left({\sqrt {2}}-1\right)}}}}{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}+1}}}}}\\&={\sqrt {2}}{\frac {\sqrt {{\sqrt[{4}]{8}}-{\sqrt {2{\sqrt {2}}-{\sqrt {2}}+1}}}}{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}+1}}}}}\\&={\sqrt {2}}{\frac {\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}+1}}}}{\sqrt {{\sqrt[{4}]{8}}-{\sqrt {{\sqrt {2}}+1}}}}}\\&={\sqrt {2}}\end{aligned}}$

 

Exercice 7-4

Simplifier les expressions littérales suivantes :

a)  ${\frac {a+{\sqrt {a^{2}-x^{2}}}}{a-{\sqrt {a^{2}-x^{2}}}}}+{\frac {a-{\sqrt {a^{2}-x^{2}}}}{a+{\sqrt {a^{2}-x^{2}}}}}$

b)  ${\frac {x+{\sqrt {x^{2}-1}}}{x-{\sqrt {x^{2}-1}}}}-{\frac {x-{\sqrt {x^{2}-1}}}{x+{\sqrt {x^{2}-1}}}}$

c)  ${\frac {\sqrt {1+x^{2}}}{\sqrt {1-x^{2}}}}+{\frac {\sqrt {1-x^{2}}}{\sqrt {1+x^{2}}}}-{\frac {2}{\sqrt {1-x^{4}}}}$

d)  ${\frac {\left(x+{\sqrt {x^{2}-a^{2}}}\right)^{4}-a^{4}}{4\left(x+{\sqrt {x^{2}-a^{2}}}\right)^{2}}}$

e)  ${\frac {x^{2}-x-2+(x-1){\sqrt {x^{2}-4}}}{x^{2}+x-2+(x+1){\sqrt {x^{2}-4}}}}$

a) Corrigé.

${\begin{aligned}{\frac {a+{\sqrt {a^{2}-x^{2}}}}{a-{\sqrt {a^{2}-x^{2}}}}}+{\frac {a-{\sqrt {a^{2}-x^{2}}}}{a+{\sqrt {a^{2}-x^{2}}}}}&={\frac {(a+{\sqrt {a^{2}-x^{2}}})(a+{\sqrt {a^{2}-x^{2}}})}{(a-{\sqrt {a^{2}-x^{2}}})(a+{\sqrt {a^{2}-x^{2}}})}}+{\frac {(a-{\sqrt {a^{2}-x^{2}}})(a-{\sqrt {a^{2}-x^{2}}})}{(a+{\sqrt {a^{2}-x^{2}}})(a-{\sqrt {a^{2}-x^{2}}})}}\\&={\frac {(a+{\sqrt {a^{2}-x^{2}}})^{2}}{a^{2}-(a^{2}-x^{2})}}+{\frac {(a-{\sqrt {a^{2}-x^{2}}})^{2}}{a^{2}-(a^{2}-x^{2})}}\\&={\frac {(a+{\sqrt {a^{2}-x^{2}}})^{2}}{x^{2}}}+{\frac {(a-{\sqrt {a^{2}-x^{2}}})^{2}}{x^{2}}}\\&={\frac {a^{2}+2a{\sqrt {a^{2}-x^{2}}}+a^{2}-x^{2}+a^{2}-2a{\sqrt {a^{2}-x^{2}}}+a^{2}-x^{2}}{x^{2}}}\\&={\frac {4a^{2}-2x^{2}}{x^{2}}}\\&={\frac {2(2a^{2}-x^{2})}{x^{2}}}\end{aligned}}$

 

b) Corrigé.

${\begin{aligned}{\frac {x+{\sqrt {x^{2}-1}}}{x-{\sqrt {x^{2}-1}}}}-{\frac {x-{\sqrt {x^{2}-1}}}{x+{\sqrt {x^{2}-1}}}}&={\frac {(x+{\sqrt {x^{2}-1}})(x+{\sqrt {x^{2}-1}})}{(x-{\sqrt {x^{2}-1}})(x+{\sqrt {x^{2}-1}})}}-{\frac {(x-{\sqrt {x^{2}-1}})(x-{\sqrt {x^{2}-1}})}{(x+{\sqrt {x^{2}-1}})(x-{\sqrt {x^{2}-1}})}}\\&={\frac {(x+{\sqrt {x^{2}-1}})^{2}}{x^{2}-(x^{2}-1)}}-{\frac {(x-{\sqrt {x^{2}-1}})^{2}}{x^{2}-(x^{2}-1)}}\\&=(x+{\sqrt {x^{2}-1}})^{2}-(x-{\sqrt {x^{2}-1}})^{2}\\&=\left(x^{2}+2x{\sqrt {x^{2}-1}}+x^{2}-1\right)-\left(x^{2}-2x{\sqrt {x^{2}-1}}+x^{2}-1\right)\\&=4x{\sqrt {x^{2}-1}}\end{aligned}}$

 

c) Corrigé.

${\begin{aligned}{\frac {\sqrt {1+x^{2}}}{\sqrt {1-x^{2}}}}+{\frac {\sqrt {1-x^{2}}}{\sqrt {1+x^{2}}}}-{\frac {2}{\sqrt {1-x^{4}}}}&={\frac {{\sqrt {1+x^{2}}}{\sqrt {1+x^{2}}}}{{\sqrt {1-x^{2}}}{\sqrt {1+x^{2}}}}}+{\frac {{\sqrt {1-x^{2}}}{\sqrt {1-x^{2}}}}{{\sqrt {1+x^{2}}}{\sqrt {1-x^{2}}}}}-{\frac {2}{\sqrt {1-x^{4}}}}\\&={\frac {1+x^{2}}{\sqrt {1-x^{4}}}}+{\frac {1-x^{2}}{\sqrt {1-x^{4}}}}-{\frac {2}{\sqrt {1-x^{4}}}}\\&={\frac {1+x^{2}+1-x^{2}-2}{\sqrt {1-x^{4}}}}\\&=0\end{aligned}}$

 

d) Corrigé.

${\begin{aligned}{\frac {\left(x+{\sqrt {x^{2}-a^{2}}}\right)^{4}-a^{4}}{4\left(x+{\sqrt {x^{2}-a^{2}}}\right)^{2}}}&={\frac {\left(x+{\sqrt {x^{2}-a^{2}}}\right)^{4}\left(x-{\sqrt {x^{2}-a^{2}}}\right)^{2}-a^{4}\left(x-{\sqrt {x^{2}-a^{2}}}\right)^{2}}{4\left(x+{\sqrt {x^{2}-a^{2}}}\right)^{2}\left(x-{\sqrt {x^{2}-a^{2}}}\right)^{2}}}\\&={\frac {\left(x+{\sqrt {x^{2}-a^{2}}}\right)^{2}\left(x^{2}-(x^{2}-a^{2})\right)^{2}-a^{4}\left(x-{\sqrt {x^{2}-a^{2}}}\right)^{2}}{4\left(x^{2}-(x^{2}-a^{2})\right)^{2}}}\\&={\frac {a^{4}\left(x+{\sqrt {x^{2}-a^{2}}}\right)^{2}-a^{4}\left(x-{\sqrt {x^{2}-a^{2}}}\right)^{2}}{4a^{4}}}\\&={\frac {\left(x+{\sqrt {x^{2}-a^{2}}}\right)^{2}-\left(x-{\sqrt {x^{2}-a^{2}}}\right)^{2}}{4}}\\&={\frac {\left(x^{2}+2x{\sqrt {x^{2}-a^{2}}}+x^{2}-a^{2}\right)-\left(x^{2}-2x{\sqrt {x^{2}-a^{2}}}+x^{2}-a^{2}\right)}{4}}\\&=x{\sqrt {x^{2}-a^{2}}}\end{aligned}}$

 

e) Corrigé.

${\frac {x^{2}-x-2+(x-1){\sqrt {x^{2}-4}}}{x^{2}+x-2+(x+1){\sqrt {x^{2}-4}}}}={\frac {(x+1)(x-2)+(x-1){\sqrt {x^{2}-4}}}{(x-1)(x+2)+(x+1){\sqrt {x^{2}-4}}}}$

L'expression sous le radical n'est définie que si $x\geqslant 2$ ou si $x\leqslant -2$.

Nous devons aussi exclure la valeur $x=-2$ car elle annule le dénominateur.

Premier cas : $x\geqslant 2$

${\begin{aligned}{\frac {x^{2}-x-2+(x-1){\sqrt {x^{2}-4}}}{x^{2}+x-2+(x+1){\sqrt {x^{2}-4}}}}&={\frac {(x+1)(x-2)+(x-1){\sqrt {(x-2)(x+2)}}}{(x-1)(x+2)+(x+1){\sqrt {(x-2)(x+2)}}}}\\&={\frac {(x+1){\sqrt {x-2}}{\sqrt {x-2}}+(x-1){\sqrt {x-2}}{\sqrt {x+2}}}{(x-1){\sqrt {x+2}}{\sqrt {x+2}}+(x+1){\sqrt {x-2}}{\sqrt {x+2}}}}\\&={\frac {{\sqrt {x-2}}\left[(x+1){\sqrt {x-2}}+(x-1){\sqrt {x+2}}\right]}{{\sqrt {x+2}}\left[(x-1){\sqrt {x+2}}+(x+1){\sqrt {x-2}}\right]}}\\&={\frac {\sqrt {x-2}}{\sqrt {x+2}}}\\&={\sqrt {\frac {x-2}{x+2}}}\end{aligned}}$

Deuxième cas : $x<-2$.

${\begin{aligned}{\frac {x^{2}-x-2+(x-1){\sqrt {x^{2}-4}}}{x^{2}+x-2+(x+1){\sqrt {x^{2}-4}}}}&={\frac {(x+1)(x-2)+(x-1){\sqrt {(2-x)(-2-x)}}}{(x-1)(x+2)+(x+1){\sqrt {(2-x)(-2-x)}}}}\\&={\frac {-(x+1){\sqrt {2-x}}{\sqrt {2-x}}+(x-1){\sqrt {2-x}}{\sqrt {-2-x}}}{-(x-1){\sqrt {-2-x}}{\sqrt {-2-x}}+(x+1){\sqrt {2-x}}{\sqrt {-2-x}}}}\\&={\frac {{\sqrt {2-x}}\left[-(x+1){\sqrt {2-x}}+(x-1){\sqrt {-2-x}}\right]}{{\sqrt {-2-x}}\left[-(x-1){\sqrt {-2-x}}+(x+1){\sqrt {2-x}}\right]}}\\&=-{\frac {\sqrt {2-x}}{\sqrt {-2-x}}}\\&=-{\sqrt {\frac {2-x}{-2-x}}}\\&=-{\sqrt {\frac {x-2}{x+2}}}\end{aligned}}$

 

Exercice 7-5

On suppose que :

$ad=bc$

en déduire que :

${\sqrt {ab}}+{\sqrt {cd}}={\sqrt {(a+c)(b+d)}}$

corrigé

Si $a=0$ ou $c=0$, on vérifie aisément que la relation est vraie.

Si $a$ et $c$ non nuls, nous avons :

$(ad=bc)\Longleftrightarrow \left({\frac {b}{a}}={\frac {d}{c}}\right)$

posons alors :

$k={\frac {b}{a}}={\frac {d}{c}}$

on alors :

${\begin{cases}b=ka\\d=kc\end{cases}}$

Nous pouvons partir de :

$b+d=b+d$

qui peut s'écrire :

${\sqrt {b^{2}}}+{\sqrt {d^{2}}}={\sqrt {(b+d)^{2}}}$

soit :

${\sqrt {kab}}+{\sqrt {kcd}}={\sqrt {(ka+kc)(b+d)}}$

et en simplifiant par ${\sqrt {k}}$, on obtient:

${\sqrt {ab}}+{\sqrt {cd}}={\sqrt {(a+c)(b+d)}}$

 

Exercice 7-6

Pour tout $a>0,\,b>0,\,a>b,$ établir la formule :

${\sqrt {2}}\left(2a+{\sqrt {a^{2}-b^{2}}}\right){\sqrt {a-{\sqrt {a^{2}-b^{2}}}}}={\sqrt {(a+b)^{3}}}-{\sqrt {(a-b)^{3}}}$

corrigé

Nous pouvons commencer par transformer le second membre en utilisant l'identité remarquable :

$A^{3}-B^{3}=(A-B)(A^{2}+AB+B^{2})$

En posant :

${\begin{cases}A={\sqrt {a+b}}\\B={\sqrt {a-b}}\end{cases}}$

On a alors :

${\begin{aligned}{\sqrt {(a+b)^{3}}}-{\sqrt {(a-b)^{3}}}&=\left({\sqrt {a+b}}\right)^{3}-\left({\sqrt {a-b}}\right)^{3}\\&=\left({\sqrt {a+b}}-{\sqrt {a-b}}\right)\left(a+b+{\sqrt {a+b}}{\sqrt {a-b}}+a-b\right)\\&=\left({\sqrt {a+b}}-{\sqrt {a-b}}\right)\left(2a+{\sqrt {a^{2}-b^{2}}}\right)\\\end{aligned}}$

En remplaçant dans :

${\sqrt {2}}\left(2a+{\sqrt {a^{2}-b^{2}}}\right){\sqrt {a-{\sqrt {a^{2}-b^{2}}}}}={\sqrt {(a+b)^{3}}}-{\sqrt {(a-b)^{3}}}$

on obtient :

${\sqrt {2}}\left(2a+{\sqrt {a^{2}-b^{2}}}\right){\sqrt {a-{\sqrt {a^{2}-b^{2}}}}}=\left({\sqrt {a+b}}-{\sqrt {a-b}}\right)\left(2a+{\sqrt {a^{2}-b^{2}}}\right)$

qui se simplifie par $\left(2a+{\sqrt {a^{2}-b^{2}}}\right)$ et on est donc ramené à démontrer la relation suivante :

${\sqrt {2}}{\sqrt {a-{\sqrt {a^{2}-b^{2}}}}}={\sqrt {a+b}}-{\sqrt {a-b}}$

qui s'écrit aussi :

${\sqrt {a-{\sqrt {a^{2}-b^{2}}}}}={\sqrt {\frac {a+b}{2}}}-{\sqrt {\frac {a-b}{2}}}$

et l'on remarque que l'on obtient directement cette relation en utilisant l'identité remarquable (vu en cours) :

${\sqrt {A-{\sqrt {B}}}}={\sqrt {\frac {A+{\sqrt {A^{2}-B}}}{2}}}-{\sqrt {\frac {A-{\sqrt {A^{2}-B}}}{2}}}$

en posant cette fois :

${\begin{cases}A=a\\B=a^{2}-b^{2}\end{cases}}$