Les fractions

Trouver, dans la liste, les nombres égaux à $-\dfrac{2}{3}$

1. $$\dfrac{-4}{6}=\dfrac{-4^{\,{-2}}}{6^{\,3}}=-\dfrac{2}{3}$$ $\Rightarrow$ oui.

2. $$\dfrac{-12}{8}=\dfrac{-4\cdot 3}{4\cdot 2}=-\dfrac{3}{2}$$ $\neq -\dfrac{2}{3}$ $\Rightarrow$ non.

3. $\dfrac{-2}{-3}=+\dfrac{2}{3}$ $\neq -\dfrac{2}{3}$ $\Rightarrow$ non.

4. $$\dfrac{10}{-15}=-\dfrac{5\cdot 2}{5\cdot 3}=-\dfrac{2}{3}$$ $\Rightarrow$ oui.

5. $\dfrac{2}{-3}=-\dfrac{2}{3}$ $\Rightarrow$ oui.

6. $$\dfrac{-2{,}1}{3{,}1}=\dfrac{-21/10}{31/10}=-\dfrac{21}{31}$$ $\neq -\dfrac{2}{3}$ $\Rightarrow$ non.

7. $$\dfrac{-1}{1{,}5}=\dfrac{-1}{\frac{3}{2}}=-\dfrac{2}{3}$$ $\Rightarrow$ oui.

Réponse : les fractions égales à $-\dfrac{2}{3}$ sont $$\displaystyle \dfrac{-4}{6},\; \dfrac{10}{-15},\; \dfrac{2}{-3},\; \dfrac{-1}{1{,}5}$$.

Écrire sous la forme d’une seule fraction

1. $$A=\dfrac{1}{5}+\dfrac{13}{5} =\dfrac{1+13}{5} =\boxed{\dfrac{14}{5}}.$$

2. $$B=\dfrac{7}{4}+\dfrac{13}{2} =\dfrac{7}{4}+\dfrac{26}{4} =\dfrac{7+26}{4} =\boxed{\dfrac{33}{4}}.$$

3. $$C=\dfrac{4}{5}+\dfrac{3}{10} =\dfrac{8}{10}+\dfrac{3}{10} =\dfrac{11}{10} =\boxed{\dfrac{11}{10}}.$$

4. $$D=\dfrac{3}{20}+\dfrac{1}{5}+\dfrac{3}{10} =\dfrac{3}{20}+\dfrac{4}{20}+\dfrac{6}{20} =\dfrac{13}{20} =\boxed{\dfrac{13}{20}}.$$

1. $$A=-\dfrac{3}{5}+\dfrac{11}{5} =\dfrac{-3+11}{5} =\boxed{\dfrac{8}{5}}.$$

2. $$B=\dfrac{7}{4}-\dfrac{13}{8} =\dfrac{14}{8}-\dfrac{13}{8} =\boxed{\dfrac{1}{8}}.$$

3. $$C=\dfrac{4}{5}+\dfrac{5}{4} =\dfrac{16}{20}+\dfrac{25}{20} =\boxed{\dfrac{41}{20}}.$$

4. $$D=-\dfrac{6}{20}+4 =-\dfrac{3}{10}+\dfrac{40}{10} =\boxed{\dfrac{37}{10}}.$$

Exprimer sous forme fractionnaire, en simplifiant

1. $$A=\dfrac{5}{6}+\dfrac{10}{6} =\dfrac{15}{6} =\dfrac{15^{\,5}}{6^{\,2}} =\boxed{\dfrac{5}{2}}.$$

2. $$B=\dfrac{12}{7}-\dfrac{4}{7} =\boxed{\dfrac{8}{7}}.$$

3. $$C=\dfrac{4}{14}+\dfrac{3}{14} =\dfrac{7}{14} =\dfrac{7}{14^{\,2}} =\boxed{\dfrac{1}{2}}.$$

4. $$D=\dfrac{5}{6}-\dfrac{1}{3} =\dfrac{5}{6}-\dfrac{2}{6} =\dfrac{3}{6} =\dfrac{3}{6^{\,2}} =\boxed{\dfrac{1}{2}}.$$

5. $$E=\dfrac{11}{10}+\dfrac{3}{8} =\dfrac{44}{40}+\dfrac{15}{40} =\boxed{\dfrac{59}{40}}.$$

6. $$F=\dfrac{4}{1}-\dfrac{6}{5} =4-\dfrac{6}{5} =\dfrac{20}{5}-\dfrac{6}{5} =\boxed{\dfrac{14}{5}}.$$

7. $$G=5+\dfrac{3}{7} =\dfrac{35}{7}+\dfrac{3}{7} =\boxed{\dfrac{38}{7}}.$$

8. $$H=\dfrac{-3}{5}-\dfrac{7}{2} =-\dfrac{6}{10}-\dfrac{35}{10} =\boxed{-\dfrac{41}{10}}.$$

9. $$I=\dfrac{5}{12}-\left(-\dfrac{7}{8}\right) =\dfrac{5}{12}+\dfrac{7}{8} =\dfrac{10}{24}+\dfrac{21}{24} =\boxed{\dfrac{31}{24}}.$$

10. $$J=\dfrac{-3}{10}+\dfrac{3}{4} =-\dfrac{6}{20}+\dfrac{15}{20} =\dfrac{9}{20} =\boxed{\dfrac{9}{20}}.$$

1. $$A=\dfrac{4}{5}\times\dfrac{7}{3} =\dfrac{4\cdot7}{5\cdot3} =\dfrac{28}{15}\,.$$

2. $$B=\dfrac{7}{5}\times\dfrac{10}{11} =\dfrac{7}{5}\times\dfrac{10^{\,2}}{11} =\dfrac{7\cdot2}{11} =\dfrac{14}{11}\,.$$

3. $$C=\dfrac{4\cdot3}{5\cdot3}+\dfrac{7\cdot5}{3\cdot5} =\dfrac{12}{15}+\dfrac{35}{15} =\dfrac{12+35}{15} =\dfrac{47}{15}\,.$$

Simplifier les quotients

1. $$A=\dfrac{\frac13}{\frac54} =\frac13\times\frac{4}{5} =\frac{4}{15}\,.$$

2. $$B=\dfrac{\frac37}{\frac75} =\frac37\times\frac{5}{7} =\frac{15}{49}\,.$$

3. $$C=-\dfrac{\frac23\times\frac45}{\frac54} =-\dfrac{\frac{8}{15}}{\frac54} =-\frac{8}{15}\times\frac{4}{5} =-\frac{32}{75}\,.$$

4. $$D=\dfrac{\frac32}{\frac{-5}{8}\times\frac47} =\dfrac{\frac32}{-\frac{20}{56}} =\dfrac{\frac32}{-\frac{5}{14}} =\frac32\times\frac{14}{-5} =\frac{42}{-10} =-\frac{21}{5}\,.$$

5. $$E=\dfrac{-\frac95\times\frac25}{\frac78\times\frac12} =\dfrac{-\frac{18}{25}}{\frac{7}{16}} =-\frac{18}{25}\times\frac{16}{7} =-\frac{288}{175}\,.$$

6. $$F=\dfrac{\frac{12}{7}\times\frac{7}{2}}{\frac57} =\dfrac{6}{\frac57} =6\times\frac{7}{5} =\frac{42}{5}\,.$$

Exprimer sous forme fractionnaire (et simplifier)

1. $$A=\dfrac45\times\dfrac73=\dfrac{4\cdot7}{5\cdot3}=\dfrac{28}{15}.$$

2. $$B=\dfrac87\times\dfrac{9}{-11} =-\dfrac{72}{77}\,.$$

3. $$\begin{aligned} C&=\frac{12}{25}\times\frac{40}{27}\\ &=\frac{480}{675}.\\[4pt] &\text{Mieux :}\\[-2pt] C&=\frac{12}{27}\times\frac{40}{25}\\ &=\frac49\times\frac85\\ &=\boxed{\frac{32}{45}}. \end{aligned}$$

4. $$D=\dfrac31\times\dfrac75=\dfrac{21}{5}.$$

5. $$E=6\times\dfrac57=\dfrac{6\cdot5}{7} =\dfrac{30}{7}.$$

6. $$F=\dfrac56\times\dfrac37 =\dfrac{15}{42} =\dfrac{15^{\,5}}{42^{\,14}} =\dfrac{5}{14}.$$

7. $$G=\dfrac{-8}{9}\times\dfrac{-3}{5} =\dfrac{24}{45} =\dfrac{24^{\,8}}{45^{\,15}} =\dfrac{8}{15}.$$

8. $$H=\dfrac{121}{63}\times\dfrac{81}{374} =\dfrac{121^{\,11}}{63}\times\dfrac{81}{374^{\,34}} =\dfrac{11\cdot 81}{63\cdot 34} =\dfrac{11\cdot 81^{\,9}}{63^{\,7}\cdot 34} =\dfrac{11\cdot 9}{7\cdot 34} =\dfrac{99}{238}\,.$$

1. $$H = 7+\dfrac{1}{3} = \dfrac{21}{3}+\dfrac{1}{3} = \dfrac{22}{3}.$$

2. $$I = 5-\dfrac{2}{3} = \dfrac{15}{3}-\dfrac{2}{3} = \dfrac{13}{3}.$$

3. $$J = -\dfrac{1}{7}+2 = -\dfrac{1}{7}+\dfrac{14}{7} = \dfrac{13}{7}.$$

4. $$K = \dfrac{1}{5}-\dfrac{1}{15} = \dfrac{3}{15}-\dfrac{1}{15} = \dfrac{2}{15}.$$

5. $$L = \dfrac{2}{9}-\dfrac{7}{18} = \dfrac{4}{18}-\dfrac{7}{18} = -\dfrac{3}{18} = -\dfrac{3}{18^{\,6}} = -\dfrac{1}{6}.$$

6. $$M = \dfrac{1}{3}+\dfrac{1}{4} = \dfrac{4}{12}+\dfrac{3}{12} = \dfrac{7}{12}.$$

7. $$N = 4\times \dfrac{7}{12} = \dfrac{4\cdot 7}{12} = \dfrac{4}{12^{\,3}}\cdot 7 = \dfrac{7}{3}.$$

8. $$O = \dfrac{1}{2}\times \dfrac{1}{6} = \dfrac{1}{12}.$$

9. $$P = \dfrac{2}{3}\times \dfrac{7}{8} = \dfrac{2\cdot 7}{3\cdot 8} = \dfrac{2^{\,1}\cdot 7}{3\cdot 8^{\,4}} = \dfrac{7}{12}.$$

10. $$Q = \dfrac{3}{4}\div 2 = \dfrac{3}{4}\times \dfrac{1}{2} = \dfrac{3}{8}.$$

11. $$R = \dfrac{1}{3}\div \dfrac{1}{2} = \dfrac{1}{3}\times \dfrac{2}{1} = \dfrac{2}{3}.$$

12. $$S = 9\div \dfrac{3}{2} = 9\times \dfrac{2}{3} = \dfrac{18}{3} = \dfrac{18^{\,6}}{3} = 6.$$

Effectuer les calculs suivants.

1. $$\dfrac{2}{7}+\dfrac{10}{7} = \dfrac{2+10}{7} = \dfrac{12}{7}\,.$$ (déjà irréductible)

2. $$\dfrac{3}{5}-\dfrac{11}{15} = \dfrac{9}{15}-\dfrac{11}{15} = \dfrac{9-11}{15} = -\dfrac{2}{15}\,.$$ (PGCD$(2,15)=1$)

3. $$\dfrac{3}{8}+\dfrac{-5}{12} = \dfrac{3}{8}-\dfrac{5}{12} = \dfrac{9}{24}-\dfrac{10}{24} = -\dfrac{1}{24}\,.$$

4. $$\dfrac{4}{5}\times\dfrac{-7}{9} = -\dfrac{4\cdot7}{5\cdot9} = -\dfrac{28}{45}\,.$$ (PGCD$(28,45)=1$)

5. $$\dfrac{-7}{8}\times\dfrac{6}{-5} = \dfrac{7}{8}\times\dfrac{6}{5} = \dfrac{7\cdot 6^{\,3}}{8^{\,4}\cdot 5} = \dfrac{7\cdot 3}{4\cdot 5} = \dfrac{21}{20}.$$

   (on a simplifié par $2$)

6. $$\dfrac{1}{2}+\left(\dfrac{1}{3}\times\dfrac{1}{4}\right) = \dfrac{1}{2}+\dfrac{1}{12} = \dfrac{6}{12}+\dfrac{1}{12} = \dfrac{7}{12}\,.$$

7. $$\begin{aligned} \frac34\times\left(\frac13-\frac25\right) &=\frac34\times\left(\frac5{15}-\frac6{15}\right)\\ &=\frac34\times\left(-\frac1{15}\right)\\ &=-\frac3{60}\\ &=\boxed{-\frac1{20}}. \end{aligned}$$

Effectuer les calculs suivants.

1. $$\dfrac{2}{5}+\dfrac{11}{5} = \dfrac{2+11}{5} = \dfrac{13}{5}\,.$$

2. $$\dfrac{7}{4}-\dfrac{11}{8} = \dfrac{14}{8}-\dfrac{11}{8} = \dfrac{3}{8}\,.$$

3. $$\dfrac{3}{7}+\dfrac{7}{3} = \dfrac{9}{21}+\dfrac{49}{21} = \dfrac{58}{21}\,.$$ (PGCD$(58,21)=1$)

4. $$\begin{aligned} \frac{40}{21}\times\frac{-70}{16} &=-\frac{40\times70}{21\times16}\\ &=-\frac52\times\frac{10}{3}\\ &=-\frac{50}{6}\\ &=\boxed{-\frac{25}{3}}. \end{aligned}$$

5. $$\begin{aligned} \frac{-700}{25}\times\frac{28}{-500} &=\frac{700}{25}\times\frac{28}{500}\\ &=28\times\frac7{125}\\ &=\boxed{\frac{196}{125}}. \end{aligned}$$

6. $$\dfrac{2}{3}+\left(\dfrac{7}{4}\times\dfrac{3}{5}\right) = \dfrac{2}{3}+\dfrac{21}{20} = \dfrac{40}{60}+\dfrac{63}{60} = \dfrac{103}{60}\,.$$ (PGCD$(103,60)=1$)

7. $$\dfrac{3}{4}\times\left(\dfrac{1}{7}-\dfrac{4}{5}\right) = \dfrac{3}{4}\times\left(\dfrac{5}{35}-\dfrac{28}{35}\right) = \dfrac{3}{4}\times\left(-\dfrac{23}{35}\right) = -\,\dfrac{69}{140}\,.$$ (PGCD$(69,140)=1$)

8. $$\dfrac{\frac{5}{7}}{\frac{35}{25}} = \dfrac{5}{7}\times\dfrac{25}{35} = \dfrac{5}{7}\times\dfrac{25^{\;5}}{35^{\;7}} = \dfrac{5}{7}\times\dfrac{5}{7} = \dfrac{25}{49}\,.$$

Exprimer sous forme de fraction, en simplifiant si possible.

1. $$\dfrac{5}{16}+\dfrac{11}{16} = \dfrac{16}{16} = 1\,.$$

2. $$\dfrac{12}{7}-\dfrac{1}{14} = \dfrac{24}{14}-\dfrac{1}{14} = \dfrac{23}{14}\,.$$ (irréductible)

3. $$\dfrac{7}{3}+\dfrac{3}{5} = \dfrac{35}{15}+\dfrac{9}{15} = \dfrac{44}{15}\,.$$

4. $$\dfrac{5}{6}-\dfrac{3}{7} = \dfrac{35}{42}-\dfrac{18}{42} = \dfrac{17}{42}\,.$$

5. $$\dfrac{11}{20}-\dfrac{7}{3} = \dfrac{33}{60}-\dfrac{140}{60} = -\,\dfrac{107}{60}\,.$$ (irréductible)

6. $$\dfrac{7}{10}+\dfrac{3}{25} = \dfrac{35}{50}+\dfrac{6}{50} = \dfrac{41}{50}\,.$$

7. $$\dfrac{5}{6}+\dfrac{1}{10} = \dfrac{25}{30}+\dfrac{3}{30} = \dfrac{28}{30} = \dfrac{28^{\;14}}{30^{\;15}} = \dfrac{14}{15}\,.$$

8. $$\dfrac{13}{20}+\dfrac{4}{15} = \dfrac{39}{60}+\dfrac{16}{60} = \dfrac{55}{60} = \dfrac{55^{\;11}}{60^{\;12}} = \dfrac{11}{12}\,.$$

Écrire chaque nombre sous forme de fraction irréductible.

1. $$\displaystyle \frac{6}{35}-\frac{2}{7}\times\frac{1}{2} = \frac{6}{35} – \frac{2\cdot 1}{7\cdot 2} = \frac{6}{35} – \frac{2\cdot 1}{7\cdot 2} = \frac{6}{35} – \frac{1}{7} = \frac{6}{35} – \frac{5}{35} = \boxed{\frac{1}{35}}.$$

2. $$\displaystyle \frac{14}{45}\times\frac{27}{49} = \frac{14^{\,2}}{45}\times\frac{27}{49^{\,7}} = \frac{2}{45^{\,5}}\times\frac{27^{\,3}}{7} = \frac{2\cdot 3}{5\cdot 7} = \boxed{\frac{6}{35}}.$$

3. $$\displaystyle \frac{9}{11}-\frac{5}{4}\div\frac{11}{7} = \frac{9}{11}-\frac{5}{4}\times\frac{7}{11} = \frac{9}{11}-\frac{35}{44} = \frac{36}{44}-\frac{35}{44} = \boxed{\frac{1}{44}}.$$

4. $$\begin{aligned} 1-\frac13-\frac14\div\frac3{16} &=1-\frac13-\frac14\times\frac{16}{3}\\ &=1-\frac13-\frac43\\ &=1-\frac53\\ &=\boxed{-\frac23}. \end{aligned}$$

5. $$\begin{aligned} \left(-\frac75+\frac43\right)\div\left(7-\frac43\right) &=\left(-\frac{21}{15}+\frac{20}{15}\right)\div\frac{17}{3}\\ &=-\frac1{15}\times\frac3{17}\\ &=-\frac3{255}\\ &=\boxed{-\frac1{85}}. \end{aligned}$$

6. $$\begin{aligned} \frac65\div\left(\frac1{15}-\frac15\right) &=\frac65\div\left(-\frac2{15}\right)\\ &=\frac65\times\left(-\frac{15}{2}\right)\\ &=\boxed{-9}. \end{aligned}$$

7. $$\begin{aligned} \frac7{18}\times\frac27-\left(\frac53-1\right)^2+1 &=\frac19-\left(\frac23\right)^2+1\\ &=\frac19-\frac49+1\\ &=-\frac13+1\\ &=\boxed{\frac23}. \end{aligned}$$

8. $$\displaystyle \frac{2\times \frac{3}{7}}{\frac{5}{3}-1} = \frac{\frac{6}{7}}{\frac{2}{3}} = \frac{6}{7}\times\frac{3}{2} = \frac{6\cdot 3}{7\cdot 2} = \frac{6^{\,3}\cdot 3}{7\cdot 2^{\,1}} = \frac{9}{7} = \boxed{\frac{9}{7}}.$$

9. $$\displaystyle \frac{\frac{11}{3}-7}{\frac{1}{9}+\frac{13}{3}} = \frac{\frac{11}{3}-\frac{21}{3}}{\frac{1}{9}+\frac{39}{9}} = \frac{-\frac{10}{3}}{\frac{40}{9}} = -\frac{10}{3}\times\frac{9}{40} = -\frac{90}{120} = -\frac{90^{\,3}}{120^{\,4}} = \boxed{-\frac{3}{4}}.$$

Équation sur une fraction

On cherche $x$ tel que la fraction obtenue en enlevant $x$ au numérateur et au dénominateur de $\dfrac{3}{7}$ vaille $5$ : $$\frac{3-x}{7-x}=5 \quad\text{avec }x\neq 7.$$ On résout : $$3-x=5(7-x)\;\Longleftrightarrow\;3-x=35-5x \;\Longleftrightarrow\;4x=32 \;\Longleftrightarrow\;x=8.$$ Vérification : $$\displaystyle \frac{3-8}{7-8}=\frac{-5}{-1}=5$$ (et $7-x=-1\neq 0$).
Réponse : $x=8$.

Décomposition en sommes d’inverses

On veut $x$ tel que $$\frac{9109}{20220}=\frac14+\frac15+\frac1x.$$ On calcule $$\frac14+\frac15=\frac{9}{20}=\frac{9099}{20220}$$ (car $20220/20=1011$). Donc $$\frac{1}{x}=\frac{9109}{20220}-\frac{9099}{20220} =\frac{10}{20220} =\frac{1}{2022}.$$ Ainsi $x=\boxed{2022}$. (Vérif : $$\,\frac14+\frac15+\frac1{2022}=\frac{9099+10}{20220}=\frac{9109}{20220}$$.)

Fraction continue simple

$$\frac{2024}{4047}=\frac{1}{\,1+\dfrac{1}{\,1+\dfrac{1}{x}\,}\,}.$$ On simplifie l’expression de droite : $$1+\frac{1}{x}=\frac{x+1}{x}\quad\Rightarrow\quad \frac{1}{\,1+\frac{1}{x}\,}=\frac{x}{x+1},$$ puis $$1+\frac{x}{x+1}=\frac{2x+1}{x+1} \quad\Rightarrow\quad \frac{1}{\,1+\dfrac{1}{\,1+\dfrac{1}{x}\,}\,} =\frac{x+1}{2x+1}.$$ On résout $$\frac{2024}{4047}=\frac{x+1}{2x+1} \;\Longleftrightarrow\; 2024(2x+1)=4047(x+1).$$ Ainsi $$4048x+2024=4047x+4047 \;\Longleftrightarrow\; x=4047-2024=\boxed{2023}.$$ (Contrôle : $(x+1)/(2x+1)=(2024)/(4047)$.)

Identité $x^3+y^3$ et application

1. Vérifications numériques. $$\frac{5^3+3^3}{5^3+2^3} =\frac{(5+3)(5^2-5\cdot3+3^2)}{(5+2)(5^2-5\cdot2+2^2)} =\frac{8\cdot19}{7\cdot19}=\frac{8}{7}=\frac{5+3}{5+2}.$$ $$\begin{aligned} \frac{13^3+10^3}{13^3+3^3} &=\frac{(13+10)(13^2-13\cdot10+10^2)} {(13+3)(13^2-13\cdot3+3^2)}\\ &=\frac{23\cdot139}{16\cdot139}\\ &=\frac{23}{16}=\frac{13+10}{13+3}. \end{aligned}$$

2. Identité. $$x^3+y^3=(x+y)(x^2-xy+y^2)\quad\text{(développement direct).}$$

3. Condition d’égalité des rapports. Soient $a,b,c\in\mathbb{N}^*$ avec $b\neq c$.

   1. Si $$\displaystyle\frac{a^3+b^3}{a^3+c^3}=\frac{a+b}{a+c}$$, en factorisant on obtient $$\frac{(a+b)\,\underline{(a^2-ab+b^2)}}{(a+c)\,\underline{(a^2-ac+c^2)}}=\frac{a+b}{a+c}.$$ Comme $a+c\neq 0$ et $a+b\neq 0$, on simplifie et il vient $$a^2-ab+b^2=a^2-ac+c^2 \;\Longleftrightarrow\; a(c-b)=c^2-b^2=(c-b)(c+b).$$ Or $c-b\neq 0$ (puisque $b\neq c$), donc $\boxed{a=b+c}$.

   2. Réciproque. Si $a=b+c$, alors $$\begin{aligned} a^2-ab+b^2 &=(b+c)^2-(b+c)b+b^2\\ &=b^2+2bc+c^2-b^2-bc+b^2\\ &=b^2+bc+c^2,\\[5pt] a^2-ac+c^2 &=(b+c)^2-(b+c)c+c^2\\ &=b^2+2bc+c^2-bc-c^2+c^2\\ &=b^2+bc+c^2. \end{aligned}$$$a^2-ab+b^2=a^2-ac+c^2$, et par le même raisonnement $$\frac{a^3+b^3}{a^3+c^3}=\frac{a+b}{a+c}.$$

4. Application. Avec $a=11$, $b=5$, $c=6$ on a $a=b+c$. Donc $$\frac{11^3+5^3}{11^3+6^3}=\frac{11+5}{11+6}=\boxed{\frac{16}{17}}.$$

Simplifier $x$, $y$ et $z$.

1. Pour $x$: $$x=\frac{\frac{\frac{\frac{\frac{4}{5}}{6}}{7}}{8}}{9} =\frac{4}{5\cdot6\cdot7\cdot8\cdot9} =\frac{4}{15120} =\boxed{\frac{1}{3780}}.$$

2. Pour $y$: le dénominateur vaut $$\frac{\frac{\frac{6}{7}}{8}}{9}=\frac{6}{7\cdot8\cdot9}=\frac{1}{84}.$$ Donc $$y=\frac{\frac{4}{5}}{\frac{1}{84}}=\frac{4}{5}\times 84=\boxed{\frac{336}{5}}.$$

3. Pour $z$: $$z=\frac{\frac{\frac{4}{5}}{6}}{\frac{\frac{7}{8}}{9}} =\frac{\frac{4}{30}}{\frac{7}{72}} =\frac{2}{15}\times\frac{72}{7} =\frac{2\cdot 72^{\,24}}{15^{\,5}\cdot 7} =\boxed{\frac{48}{35}}.$$

Posons $T=\dfrac{x+y}{x-y}$ (bien défini et $T>0$ car $x>y>0$). Alors $$T^2=\frac{(x+y)^2}{(x-y)^2} =\frac{x^2+2xy+y^2}{x^2-2xy+y^2} =\frac{(x^2+y^2)+2xy}{(x^2+y^2)-2xy}.$$ Comme $x^2+y^2=6xy$, on obtient $$T^2=\frac{6xy+2xy}{6xy-2xy}=\frac{8xy}{4xy}=2.$$ Donc $T=\sqrt{2}$ (et non $-\sqrt2$ puisque $T>0$).

$$\boxed{\displaystyle \frac{x+y}{x-y}=\sqrt{2}.}$$

2) Résoudre $$\displaystyle \frac{x}{a}+\frac{a}{b}=\frac{x}{b}+\frac{b}{a}$$, $a,b\neq0$.

$$\frac{x}{a}-\frac{x}{b}=\frac{b}{a}-\frac{a}{b} \quad\Longleftrightarrow\quad x\left(\frac{1}{a}-\frac{1}{b}\right)=\frac{b^2-a^2}{ab} =\frac{(b-a)(b+a)}{ab}.$$$$\displaystyle \frac{1}{a}-\frac{1}{b}=\frac{b-a}{ab}$$. Donc $$x\cdot\frac{b-a}{ab}=\frac{b-a}{ab}\,(a+b).$$

• Si $a\neq b$, on simplifie par $b-a\neq0$ et $\,ab\neq0$ : $\;\boxed{x=a+b}.$

• Si $a=b$, l’égalité est une identité pour $\boxed{x\in\mathbb{R}}$.

3) Résoudre $$\displaystyle \frac{x}{a}+\frac{a}{x}=\frac{x}{b}+\frac{b}{x}$$, $a,b\neq0$.

On regroupe : $$\frac{x}{a}-\frac{x}{b}=\frac{b}{x}-\frac{a}{x} \;\Longleftrightarrow\; x\left(\frac{1}{a}-\frac{1}{b}\right)=\frac{b-a}{x}.$$$x$ (avec $x\neq0$) : $$x^2\left(\frac{1}{a}-\frac{1}{b}\right)=b-a \quad\text{et}\quad \frac{1}{a}-\frac{1}{b}=\frac{b-a}{ab}.$$$$x^2\cdot\frac{b-a}{ab}=b-a.$$

• Si $a\neq b$, on simplifie par $b-a\neq0$ : $$\displaystyle \frac{x^2}{ab}=1\Rightarrow x^2=ab$$.

  • Si $ab>0$ : $\boxed{x=\pm\sqrt{ab}}$.

  • Si $ab<0$ : $\boxed{\text{pas de solution réelle}}$.

• Si $a=b$, l’égalité est vraie pour tout $\boxed{x\neq0}$ (car les deux membres sont égaux).

4) Moyenne pondérée de Laure (SVT).

Les notes et coefficients donnent la somme pondérée $$14\times3+17\times2+18\times1+16\times2+ x\times2=126+2x,$$ pour un total de coefficients $3+2+1+2+2=10$. L’équation de la moyenne $16$ est bien $\displaystyle \frac{126+2x}{10}=16$.

1. C’est justifié ci-dessus : $\displaystyle \frac{126+2x}{10}=16$.

2. Résolution : $$126+2x=160\Rightarrow 2x=34\Rightarrow \boxed{x=17}$$, admissible ($0\le x\le20$).

5) Simplifier avec les puissances (sans calculatrice). $$\bigl(2=2^1,\;3=3^1,\;6=2\cdot3,\;8=2^3,\;16=2^4,\;24=2^3\cdot3,\;32=2^5\bigr)$$