Corrigé Exercice 20 : Ensemble $\mathbb{Q}$ des nombres rationnels 4e

 

Calculons puis rendons irréductible.

 

Soit : $A=\dfrac{\dfrac{1}{3}}{\dfrac{2}{7}\times\dfrac{1}{4}}+\dfrac{\dfrac{\dfrac{1}{2}}{5}\times\dfrac{1}{8}}{\dfrac{1}{7}\div\dfrac{3}{\dfrac{5}{2}\times 4}}$

 

Alors, en calculant, on trouve :

 

$\begin{array}{rcl} A&=&\dfrac{\dfrac{1}{3}}{\dfrac{2}{7}\times\dfrac{1}{4}}+\dfrac{\dfrac{\dfrac{1}{2}}{5}\times\dfrac{1}{8}}{\dfrac{1}{7}\div\dfrac{3}{\dfrac{5}{2}\times 4}}\\\\&=&\dfrac{\dfrac{1}{3}}{\dfrac{2}{28}}+\dfrac{\dfrac{1}{2}\times\dfrac{1}{5}\times\dfrac{1}{8}}{\dfrac{1}{7}\div\dfrac{3}{\dfrac{20}{2}}}\\\\&=&\dfrac{1}{3}\times\dfrac{28}{2}+\dfrac{\dfrac{1}{80}}{\dfrac{1}{7}\div\dfrac{3}{10}}\\\\&=&\dfrac{28}{6}+\dfrac{\dfrac{1}{80}}{\dfrac{1}{7}\times\dfrac{10}{3}}\\\\&=&\dfrac{14}{3}+\dfrac{\dfrac{1}{80}}{\dfrac{10}{21}}\\\\&=&\dfrac{14}{3}+\left(\dfrac{1}{80}\times\dfrac{21}{10}\right)\\\\&=&\dfrac{14}{3}+\dfrac{21}{800}\\\\&=&\dfrac{11\,200}{24\,000}+\dfrac{63}{24\,000}\\\\&=&\dfrac{11\,263}{24\,000}\end{array}$

 

D'où, $\boxed{A=\dfrac{11\,263}{24\,000}}$

 

Soit : $B=\dfrac{(-2)^{2}\times\dfrac{5}{3}}{7-\dfrac{2}{3}}\div\dfrac{(-1)^{9}+\dfrac{4}{9}}{1-\dfrac{2}{11}}$

 

On rappelle que si $n$ est un entier naturel alors :

 

$\ \centerdot\ (-1)^{n}=1$ si $n$ est pair ; c'est-à-dire, si $n$ est un multiple de $2$

 

$\ \centerdot\ (-1)^{n}=-1$ si $n$ est impair ; c'est-à-dire, si $n$ n'est pas multiple de $2$

 

Alors, en calculant, on trouve :

 

$\begin{array}{rcl} B&=&\dfrac{(-2)^{2}\times\dfrac{5}{3}}{7-\dfrac{2}{3}}\div\dfrac{(-1)^{9}+\dfrac{4}{9}}{1-\dfrac{2}{11}}\\\\&=&\dfrac{4\times\dfrac{5}{3}}{\dfrac{21}{3}-\dfrac{2}{3}}\div\dfrac{-1+\dfrac{4}{9}}{\dfrac{11}{11}-\dfrac{2}{11}}\\\\&=&\dfrac{\dfrac{20}{3}}{\dfrac{21-2}{3}}\div\dfrac{\dfrac{-9}{9}+\dfrac{4}{9}}{\dfrac{11-2}{11}}\\\\&=&\dfrac{\dfrac{20}{3}}{\dfrac{19}{3}}\div\dfrac{\dfrac{-9+4}{9}}{\dfrac{9}{11}}\\\\&=&\left(\dfrac{20}{3}\times\dfrac{3}{19}\right)\div\dfrac{\dfrac{-5}{9}}{\dfrac{9}{11}}\\\\&=&\dfrac{20}{19}\div\left(\dfrac{-5}{9}\times\dfrac{11}{9}\right)\\\\&=&\dfrac{20}{19}\div\left(\dfrac{-55}{81}\right)\\\\&=&\dfrac{20}{19}\times\dfrac{81}{-55}\\\\&=&\dfrac{4}{19}\times\dfrac{81}{-11}\\\\&=&\dfrac{324}{-209}\end{array}$

 

Donc, $\boxed{B=-\dfrac{324}{209}}$

 

Soit : $C=\dfrac{\dfrac{1}{3}}{\dfrac{2}{7}\times\dfrac{1}{4}}-\dfrac{\dfrac{4}{5}\times\dfrac{1}{8}}{\dfrac{1}{7}\times\dfrac{3}{4}}$

 

Alors, en calculant, on trouve :

 

$\begin{array}{rcl} C&=&\dfrac{\dfrac{1}{3}}{\dfrac{2}{7}\times\dfrac{1}{4}}-\dfrac{\dfrac{4}{5}\times\dfrac{1}{8}}{\dfrac{1}{7}\times\dfrac{3}{4}}\\\\&=&\dfrac{\dfrac{1}{3}}{\dfrac{2}{28}}-\dfrac{\dfrac{4}{40}}{\dfrac{3}{28}}\\\\&=&\left(\dfrac{1}{3}\times\dfrac{28}{2}\right)-\left(\dfrac{4}{40}\times\dfrac{28}{3}\right)\\\\&=&\left(\dfrac{1}{3}\times\dfrac{14}{1}\right)-\left(\dfrac{1}{10}\times\dfrac{28}{3}\right)\\\\&=&\dfrac{14}{3}-\dfrac{14}{15}\\\\&=&\dfrac{70}{15}-\dfrac{14}{15}\\\\&=&\dfrac{56}{15}\end{array}$

 

Ainsi, $\boxed{C=\dfrac{56}{15}}$

 

Soit : $D=\dfrac{\dfrac{1}{3}}{\dfrac{2}{7}+\dfrac{1}{4}}\times\dfrac{\dfrac{\dfrac{1}{4}}{5}-\dfrac{1}{8}}{\dfrac{1}{7}-\dfrac{3}{\dfrac{5}{2}+4}}$

 

Alors, en calculant, on obtient :

 

$\begin{array}{rcl} D&=&\dfrac{\dfrac{1}{3}}{\dfrac{2}{7}+\dfrac{1}{4}}\times\dfrac{\dfrac{\dfrac{1}{4}}{5}-\dfrac{1}{8}}{\dfrac{1}{7}-\dfrac{3}{\dfrac{5}{2}+4}}\\\\&=&\dfrac{\dfrac{1}{3}}{\dfrac{8}{28}+\dfrac{7}{28}}\times\dfrac{\dfrac{1}{4}\times\dfrac{1}{5}-\dfrac{1}{8}}{\dfrac{1}{7}-\dfrac{3}{\dfrac{5}{2}+\dfrac{8}{2}}}\\\\&=&\dfrac{\dfrac{1}{3}}{\dfrac{15}{28}}\times\dfrac{\dfrac{1}{20}-\dfrac{1}{8}}{\dfrac{1}{7}-\dfrac{3}{\dfrac{13}{2}}}\\\\&=&\dfrac{1}{3}\times\dfrac{28}{15}\times\dfrac{\dfrac{8}{160}-\dfrac{20}{160}}{\dfrac{1}{7}-\dfrac{3}{1}\times\dfrac{2}{13}}\\\\&=&\dfrac{28}{45}\times\dfrac{\dfrac{-12}{160}}{\dfrac{1}{7}-\dfrac{6}{13}}\\\\&=&\dfrac{28}{45}\times\dfrac{\dfrac{-12}{160}}{\dfrac{13}{91}-\dfrac{42}{91}}\\\\&=&\dfrac{28}{45}\times\dfrac{\dfrac{-12}{160}}{\dfrac{-29}{91}}\\\\&=&\dfrac{28}{45}\times\left(\dfrac{-12}{160}\times\dfrac{91}{-29}\right)\\\\&=&\dfrac{28}{45}\times\dfrac{12}{160}\times\dfrac{91}{29}\\\\&=&\dfrac{4\times 7\times 3\times 4\times 91}{3\times 15\times 4\times 4\times 10\times 29}\\\\&=&\dfrac{7\times 91}{15\times 10\times 29}\\\\&=&\dfrac{637}{4\,350}\end{array}$

 

D'où, $\boxed{D=\dfrac{637}{4\,350}}$