From c1143129416cf7211e5b3b2341d63dc502ee0abc Mon Sep 17 00:00:00 2001 From: Bertrand Benjamin Date: Tue, 4 Nov 2025 09:29:57 +0100 Subject: [PATCH] feat(2nd): add correction --- 2nd/02_Calcul_Litteral/1_exercises.tex | 4 +- 2nd/02_Calcul_Litteral/exercises.tex | 277 +++++++++++++++++++++++++ 2nd/02_Calcul_Litteral/solutions.pdf | Bin 21736 -> 47492 bytes 3 files changed, 279 insertions(+), 2 deletions(-) diff --git a/2nd/02_Calcul_Litteral/1_exercises.tex b/2nd/02_Calcul_Litteral/1_exercises.tex index 8413f6c..17cc18e 100644 --- a/2nd/02_Calcul_Litteral/1_exercises.tex +++ b/2nd/02_Calcul_Litteral/1_exercises.tex @@ -41,7 +41,7 @@ \end{align*} \item \begin{align*} - F & = \dfrac{- 9}{9} + 9a + 4a + 8 \\ & = 9a + \dfrac{- 9}{9} + 4a + 8 \\ & = 9a + 4a + \dfrac{- 9}{9} + 8 \\ & = (9 + 4) \times a + \dfrac{- 9}{9} + \dfrac{8}{1} \\ & = 13a + \dfrac{- 9}{9} + \dfrac{8 \times 9}{1 \times 9} \\ & = 13a + \dfrac{- 9}{9} + \dfrac{72}{9} \\ & = 13a + \dfrac{- 9}{9} + \dfrac{72}{9} \\ & = 13a + \dfrac{- 9 + 72}{9} \\ & = 13a + \dfrac{63}{9} + F & = \dfrac{- 9}{9} + 9a + 4a + 8 \\ & = 9a + \dfrac{- 9}{9} + 4a + 8 \\ & = 9a + 4a + \dfrac{- 9}{9} + 8 \\ & = (9 + 4) \times a + \dfrac{- 9}{9} + \dfrac{8}{1} \\ & = 13a + \dfrac{- 9}{9} + \dfrac{8 \times 9}{1 \times 9} \\ & = 13a + \dfrac{- 9}{9} + \dfrac{72}{9} \\ & = 13a + \dfrac{- 9}{9} + \dfrac{72}{9} \\ & = 13a + \dfrac{- 9 + 72}{9} \\ & = 13a + \dfrac{63}{9} \\ & = 13a + 7 \end{align*} \item \begin{align*} @@ -99,7 +99,7 @@ \end{align*} \item \begin{align*} - F & = \dfrac{2}{10} \times x(2x + 9) \\ & = \dfrac{2}{10} \times x \times 2x + \dfrac{2}{10} \times x \times 9 \\ & = \dfrac{2}{10} \times 2 \times x^{1 + 1} + 9 \times \dfrac{2}{10} \times x \\ & = \dfrac{2 \times 2}{10} \times x^{2} + \dfrac{9 \times 2}{10} \times x \\ & = \dfrac{4}{10} \times x^{2} + \dfrac{18}{10} \times x + F & = \dfrac{2}{10} \times x(2x + 9) \\ & = \dfrac{2}{10} \times x \times 2x + \dfrac{2}{10} \times x \times 9 \\ & = \dfrac{2}{10} \times 2 \times x^{1 + 1} + 9 \times \dfrac{2}{10} \times x \\ & = \dfrac{2 \times 2}{10} \times x^{2} + \dfrac{9 \times 2}{10} \times x \\ & = \dfrac{4}{10} \times x^{2} + \dfrac{18}{10} \times x \\ & = \dfrac{2}{5} x^{2} + \dfrac{9}{5} x \end{align*} \end{enumerate} \end{multicols} diff --git a/2nd/02_Calcul_Litteral/exercises.tex b/2nd/02_Calcul_Litteral/exercises.tex index 5bc15b5..19c89c5 100644 --- a/2nd/02_Calcul_Litteral/exercises.tex +++ b/2nd/02_Calcul_Litteral/exercises.tex @@ -30,6 +30,31 @@ Qu'en pensez-vous ? \end{exercise} +\begin{solution} + Pour vérifier si Bob a raison, on va traduire chaque programme par une expression littérale en notant $x$ le nombre de départ. + + \textbf{Programme A :} + \begin{itemize} + \item Choisir un nombre : $x$ + \item Multiplier par 4 : $4x$ + \item Soustraire 1 : $4x - 1$ + \item Ajouter le nombre de départ : $4x - 1 + x$ + \item Soustraire 2 : $4x - 1 + x - 2$ + \end{itemize} + On réduit l'expression : $4x - 1 + x - 2 = 5x - 3$ + + \textbf{Programme B :} + \begin{itemize} + \item Choisir un nombre : $x$ + \item Multiplier par 5 : $5x$ + \item Enlever 3 : $5x - 3$ + \end{itemize} + + Les deux programmes donnent bien la même expression finale : $5x - 3$. + + \textbf{Conclusion :} Bob a raison, ces deux programmes donnent toujours le même résultat. +\end{solution} + \begin{exercise}[subtitle={Vrai ou faux}, step={1}, origin={MEpC}, topics={ Fraction Developpement Litteral }, tags={ Fractions, Developpement }, mode={\searchMode}] Pour chacune des affirmations, expliquer si elles sont vraies ou fausses. \begin{enumerate} @@ -41,6 +66,34 @@ \end{enumerate} \end{exercise} +\begin{solution} + \begin{enumerate} + \item \textbf{Faux.} On ne peut pas additionner un nombre et un terme en $x$. Par exemple, pour $x = 1$ : $4 + 3 \times 1 = 7$ mais $7 \times 1 = 7$. Cependant, pour $x = 2$ : $4 + 3 \times 2 = 10$ mais $7 \times 2 = 14$. L'égalité n'est donc pas vraie pour tous les nombres. + + \item \textbf{Faux.} Cette égalité n'est vraie que pour $y = 0$ et $y = 1$. Par exemple, pour $y = 2$ : $2^2 = 4 \neq 2$. + + \item \textbf{Vrai.} Réduisons chaque membre : + \begin{itemize} + \item Membre de gauche : $2z + z - 8 = 3z - 8$ + \item Membre de droite : $3z - 7 - 1 = 3z - 8$ + \end{itemize} + Les deux expressions sont égales. + + \item \textbf{Faux.} On ne peut pas simplifier ainsi. En développant le membre de gauche : + \[ + \dfrac{4t-8}{8} = \dfrac{4t}{8} - \dfrac{8}{8} = \dfrac{t}{2} - 1 + \] + Ce qui est différent de $4t - 1$. + + \item \textbf{Faux.} Réduisons chaque membre : + \begin{itemize} + \item Membre de gauche : $3t + 3 + 5 = 3t + 8$ + \item Membre de droite : $t + 2t + 4 = 3t + 4$ + \end{itemize} + Les deux expressions ne sont pas égales ($8 \neq 4$). + \end{enumerate} +\end{solution} + \begin{exercise}[subtitle={Aire de rectangles}, step={2}, origin={Classique}, topics={ Fraction Developpement Litteral }, tags={ Fractions, Developpement }, mode={\searchMode}] Trouver deux façons différentes de calculer l'aire de ces rectangles. @@ -100,6 +153,36 @@ \end{multicols} \end{exercise} +\begin{solution} + Pour chaque rectangle, on peut calculer l'aire de deux façons différentes : soit en calculant l'aire totale directement, soit en découpant le rectangle et en additionnant les aires des parties. + + \begin{enumerate} + \item \textbf{Méthode 1 :} Aire totale = $(1 + x) \times 3 = 3 + 3x$ + + \textbf{Méthode 2 :} Somme des aires = $1 \times 3 + x \times 3 = 3 + 3x$ + + On obtient bien la même expression : $3 + 3x = 3(1 + x)$ + + \item \textbf{Méthode 1 :} Aire totale = $4 \times (2 + x) = 8 + 4x$ + + \textbf{Méthode 2 :} Somme des aires = $4 \times 2 + 4 \times x = 8 + 4x$ + + On obtient bien la même expression : $8 + 4x = 4(2 + x)$ + + \item \textbf{Méthode 1 :} Aire totale = $(x + 1) \times (3 + x) = 3x + x^2 + 3 + x = x^2 + 4x + 3$ + + \textbf{Méthode 2 :} Somme des aires = $x \times 3 + x \times x + 1 \times 3 + 1 \times x = 3x + x^2 + 3 + x = x^2 + 4x + 3$ + + On obtient bien la même expression. + + \item \textbf{Méthode 1 :} Aire totale = $(6x + 3) \times (2 + 2x) = 12x + 12x^2 + 6 + 6x = 12x^2 + 18x + 6$ + + \textbf{Méthode 2 :} Somme des aires = $6x \times 2 + 6x \times 2x + 3 \times 2 + 3 \times 2x = 12x + 12x^2 + 6 + 6x = 12x^2 + 18x + 6$ + + On obtient bien la même expression. + \end{enumerate} +\end{solution} + \begin{exercise}[subtitle={Masse volumique}, step={3}, origin={Classique}, topics={ Fraction Developpement Litteral }, tags={ Fractions, Developpement }, mode={\trainMode}] La masse volumique $\rho$ (prononcer “rhô”) d’un échantillon de matière est une grandeur qui caractérise une espèce chimique. Elle dépend de son état (solide, liquide ou gaz) et de la température ambiante. Elle s’exprime en $g/L$. Elle est égale au quotient de sa masse $m$ (en g) par le volume $V$ (en L) qu’il occupe : \[ @@ -108,6 +191,16 @@ Sachant que la masse de $10 L$ d’acétone est de $\np{7840} g$, quelle est sa masse volumique ? \end{exercise} +\begin{solution} + On utilise la formule $\rho = \dfrac{m}{V}$ avec $m = \np{7840}\,g$ et $V = 10\,L$. + + \[ + \rho = \dfrac{\np{7840}}{10} = 784\,g/L + \] + + La masse volumique de l'acétone est de $784\,g/L$. +\end{solution} + \begin{exercise}[subtitle={Avec formulaire}, step={3}, origin={Classique}, topics={ Fraction Developpement Litteral }, tags={ Fractions, Developpement }, mode={\trainMode}] À l'aide du formulaire fourni, et sans calculatrice, calculer les grandeurs suivantes en s'assurant de respecter les unités du Système International données par le formulaire : \begin{enumerate} @@ -120,6 +213,7 @@ \item L’énergie électrique consommée en 60 s par un radiateur dont la puissance est de $2000 W$. \item La force de gravitation qu’exerce un potiron de $2 kg$ sur une citrouille de $4 kg$ situés à une distance de $4 m$ l’un de l’autre \item La masse volumique de l'éthanol, à partir d'un échantillon de $2\,L$ d'éthanol de masse $1,578\,kg$ + \item L’énergie cinétique d’une prune d’une masse de $8 g$ tombant d’un arbre à une vitesse de $2 m/s$ \item La vitesse moyenne d’un vélo ayant parcouru $22 km$ en 1 h et 6 min \item La quantité de matière équivalant à $602 \times 10^{23}$ atomes d'oxygène @@ -131,6 +225,77 @@ \end{enumerate} \end{exercise} +\begin{solution} + \begin{enumerate} + \item Formule : $U = R \times I$ avec $R = 4\,\Omega$ et $I = 2,5\,A$ + + $U = 4 \times 2,5 = 10\,V$ + + \item Formule : $T = \dfrac{1}{f}$ avec $f = 5\,Hz$ + + $T = \dfrac{1}{5} = 0,2\,s$ + + \item Formule : $C_m = \dfrac{m}{V}$ avec $m = 8\,g$ et $V = 4\,L$ + + $C_m = \dfrac{8}{4} = 2\,g/L$ + + \item Formule : $E_c = \dfrac{1}{2}mv^2$ avec $m = 0,1\,kg$ et $v = 2\,m/s$ + + $E_c = \dfrac{1}{2} \times 0,1 \times 2^2 = \dfrac{1}{2} \times 0,1 \times 4 = 0,2\,J$ + + \item Formule : $v = \dfrac{d}{t}$ avec $d = 35\,m$ et $t = 7\,s$ + + $v = \dfrac{35}{7} = 5\,m/s$ + + \item Formule : $P = m \times g$ avec $m = 1000\,kg$ et $g = 9,8\,N/kg$ + + $P = 1000 \times 9,8 = \np{9800}\,N$ + + \item Formule : $E = P \times t$ avec $P = 2000\,W$ et $t = 60\,s$ + + $E = 2000 \times 60 = \np{120000}\,J$ + + \item Formule : $F = G \times \dfrac{m_1 \times m_2}{d^2}$ avec $G = 6,67 \times 10^{-11}\,N \cdot m^2/kg^2$, $m_1 = 2\,kg$, $m_2 = 4\,kg$ et $d = 4\,m$ + + $F = 6,67 \times 10^{-11} \times \dfrac{2 \times 4}{4^2} = 6,67 \times 10^{-11} \times \dfrac{8}{16} = 6,67 \times 10^{-11} \times 0,5 = 3,335 \times 10^{-11}\,N$ + + \item Formule : $\rho = \dfrac{m}{V}$ avec $m = 1,578\,kg = 1578\,g$ et $V = 2\,L$ + + $\rho = \dfrac{1578}{2} = 789\,g/L$ + + \item Formule : $E_c = \dfrac{1}{2}mv^2$ avec $m = 8\,g = 0,008\,kg$ et $v = 2\,m/s$ + + $E_c = \dfrac{1}{2} \times 0,008 \times 2^2 = \dfrac{1}{2} \times 0,008 \times 4 = 0,016\,J$ + + \item Formule : $v = \dfrac{d}{t}$ avec $d = 22\,km = \np{22000}\,m$ et $t = 1\,h\,6\,min = 66\,min = 3960\,s$ + + $v = \dfrac{\np{22000}}{3960} \approx 5,56\,m/s$ + + \item Formule : $n = \dfrac{N}{N_A}$ avec $N = 602 \times 10^{23}$ et $N_A = 6,02 \times 10^{23}\,mol^{-1}$ + + $n = \dfrac{602 \times 10^{23}}{6,02 \times 10^{23}} = 100\,mol$ + + \item Formule : $P = m \times g$ avec $m = 400\,t = \np{400000}\,kg$ et $g = 9,8\,N/kg$ + + $P = \np{400000} \times 9,8 = \np{3920000}\,N$ + + \item Formule : $U = R \times I$ avec $R = 4\,c\Omega = 0,04\,\Omega$ et $I = 2,5\,mA = 0,0025\,A$ + + $U = 0,04 \times 0,0025 = 0,0001\,V = 0,1\,mV$ + + \item Formule : $E = P \times t$ avec $P = 3\,kW = 3000\,W$ et $t = 24\,h \times 0,3 = 7,2\,h = 25920\,s$ + + $E = 3000 \times 25920 = \np{77760000}\,J$ + + \item Formule : $C_m = \dfrac{m}{V}$ avec $m = 0,2\,kg = 200\,g$ et $V = 0,1\,m^3 = 100\,L$ + + $C_m = \dfrac{200}{100} = 2\,g/L$ + + \item Formule : $F = G \times \dfrac{m_1 \times m_2}{d^2}$ avec $G = 6,67 \times 10^{-11}\,N \cdot m^2/kg^2$, $m_1 = 500\,g = 0,5\,kg$, $m_2 = 100\,g = 0,1\,kg$ et $d = 10\,cm = 0,1\,m$ + + $F = 6,67 \times 10^{-11} \times \dfrac{0,5 \times 0,1}{0,1^2} = 6,67 \times 10^{-11} \times \dfrac{0,05}{0,01} = 6,67 \times 10^{-11} \times 5 = 3,335 \times 10^{-10}\,N$ + \end{enumerate} +\end{solution} \begin{exercise}[subtitle={Transformation du formulaire}, step={3}, origin={Classique}, topics={ Fraction Developpement Litteral }, tags={ Fractions, Developpement }, mode={\trainMode}] À l'aide du formulaire fourni, calculer les grandeurs suivantes en s'assurant de respecter les unités du Système International données par le formulaire. On commencera par manipuler la formule afin d'isoler la grandeur recherchée, avant de faire l'application numérique : @@ -145,6 +310,66 @@ \end{enumerate} \end{exercise} +\begin{solution} + \begin{enumerate} + \item On cherche $R$ dans la formule $U = R \times I$ + + On isole $R$ : $R = \dfrac{U}{I}$ + + Application numérique avec $U = 4\,V$ et $I = 160\,mA = 0,16\,A$ : + + $R = \dfrac{4}{0,16} = 25\,\Omega$ + + \item On cherche $f$ dans la formule $T = \dfrac{1}{f}$ + + On isole $f$ : $f = \dfrac{1}{T}$ + + Application numérique avec $T = 0,5\,s$ : + + $f = \dfrac{1}{0,5} = 2\,Hz$ + + \item On cherche $m$ dans la formule $C_m = \dfrac{m}{V}$ + + On isole $m$ : $m = C_m \times V$ + + Application numérique avec $C_m = 9\,g/L$ et $V = 2,5\,L$ : + + $m = 9 \times 2,5 = 22,5\,g$ + + \item On cherche $m$ dans la formule $E_c = \dfrac{1}{2}mv^2$ + + On isole $m$ : $m = \dfrac{2E_c}{v^2}$ + + Application numérique avec $E_c = 1\,J$ et $v = 2\,m/s$ : + + $m = \dfrac{2 \times 1}{2^2} = \dfrac{2}{4} = 0,5\,kg$ + + \item On cherche $m_2$ dans la formule $F = G \times \dfrac{m_1 \times m_2}{d^2}$ + + On isole $m_2$ : $m_2 = \dfrac{F \times d^2}{G \times m_1}$ + + Application numérique avec $F = 6,67 \times 10^{-11}\,N$, $d = 1\,m$, $m_1 = 2000\,g = 2\,kg$ et $G = 6,67 \times 10^{-11}\,N \cdot m^2/kg^2$ : + + $m_2 = \dfrac{6,67 \times 10^{-11} \times 1^2}{6,67 \times 10^{-11} \times 2} = \dfrac{6,67 \times 10^{-11}}{13,34 \times 10^{-11}} = 0,5\,kg$ + + \item On cherche $P$ dans la formule $E = P \times t$ + + On isole $P$ : $P = \dfrac{E}{t}$ + + Application numérique avec $E = 2400\,J$ et $t = 60\,s$ : + + $P = \dfrac{2400}{60} = 40\,W$ + + \item On cherche $v$ dans la formule $E_c = \dfrac{1}{2}mv^2$ + + On isole $v$ : $v^2 = \dfrac{2E_c}{m}$ donc $v = \sqrt{\dfrac{2E_c}{m}}$ + + Application numérique avec $E_c = 10\,mJ = 0,01\,J$ et $m = 5\,g = 0,005\,kg$ : + + $v = \sqrt{\dfrac{2 \times 0,01}{0,005}} = \sqrt{\dfrac{0,02}{0,005}} = \sqrt{4} = 2\,m/s$ + \end{enumerate} +\end{solution} + \begin{exercise}[subtitle={Emballage}, step={3}, origin={Classique}, topics={ Fraction Developpement Litteral }, tags={ Fractions, Developpement }, mode={\searchMode}] \begin{enumerate} @@ -154,3 +379,55 @@ \item On considère un parallélépipède rectangle de longueur $L$, de largeur $l$ et de hauteur $h$. On note son volume $V$ et l'aire totale de ses faces $A$. Noter toutes les formules que vous avez utilisées pour répondre aux questions précédentes. \end{enumerate} \end{exercise} + +\begin{solution} + \begin{enumerate} + \item Pour calculer la hauteur de la boîte, on utilise la formule du volume : + + $V = L \times l \times h$ donc $h = \dfrac{V}{L \times l}$ + + Avec $V = 1\,L = 1000\,cm^3$, $L = 15\,cm$ et $l = 8\,cm$ : + + $h = \dfrac{1000}{15 \times 8} = \dfrac{1000}{120} \approx 8,33\,cm$ + + La hauteur de la boîte est d'environ $8,33\,cm$, ce qui est bien inférieur à $20\,cm$. \textbf{La boîte rentrera donc dans le frigo.} + + \item Pour emballer le livre, on doit couvrir toutes les faces visibles. Un livre a 6 faces, mais une face (celle qui repose) n'est généralement pas emballée. On emballe donc 5 faces. + + Les dimensions du livre sont : longueur $L = 20\,cm$, largeur $l = 15\,cm$, épaisseur $h = 3\,cm$. + + Surface d'emballage visible : + \begin{itemize} + \item Face avant et arrière : $2 \times (L \times h) = 2 \times (20 \times 3) = 120\,cm^2$ + \item Côtés gauche et droit : $2 \times (l \times h) = 2 \times (15 \times 3) = 90\,cm^2$ + \item Face supérieure : $L \times l = 20 \times 15 = 300\,cm^2$ + \end{itemize} + + Surface totale : $120 + 90 + 300 = 510\,cm^2 = 0,051\,m^2$ + + \item On cherche l'épaisseur maximale $h$ sachant que la surface d'emballage ne doit pas dépasser $0,12\,m^2 = 1200\,cm^2$. + + Avec les mêmes dimensions $L = 20\,cm$ et $l = 15\,cm$, on a : + + $A = 2Lh + 2lh + Ll = 2h(L + l) + Ll$ + + On isole $h$ : + + $2h(L + l) = A - Ll$ + + $h = \dfrac{A - Ll}{2(L + l)}$ + + Application numérique : + + $h = \dfrac{1200 - 20 \times 15}{2(20 + 15)} = \dfrac{1200 - 300}{2 \times 35} = \dfrac{900}{70} \approx 12,86\,cm$ + + L'épaisseur maximale est d'environ $12,86\,cm$. + + \item Formules utilisées : + \begin{itemize} + \item Volume d'un parallélépipède rectangle : $V = L \times l \times h$ + \item Aire totale des faces : $A = 2(Ll + Lh + lh)$ + \item Aire de 5 faces (sans la base) : $A = 2Lh + 2lh + Ll$ + \end{itemize} + \end{enumerate} +\end{solution} diff --git a/2nd/02_Calcul_Litteral/solutions.pdf b/2nd/02_Calcul_Litteral/solutions.pdf index 1a7b03fd208afa43b5bd5b0fdd9db02b580d1714..79fca7d889f1df9e0aac27827522316ae35c9b89 100644 GIT binary patch literal 47492 zcmce7V~{AplJ3~HJ#)sk&)Bx-jBVStZQHhO+qV7Ay}KK4BX(o|-P;jeRh`vc-Ct$q zS5+C+q;f(cGz_$CP^6PfeTze@vxCD>%mnlVHh;{axVZ`Fl-z8M2^66b50PrusTPP!I!~d-LTl&ZMv;A%P--m`2m|B44#~gQwiy0(#>My#b^rTx!uSsp_}5q$Bh$YZ$vzY}3?%>Klc7QOz?sd2XEop8o z1SOM(v|n7-Wp~^;WZR1Mfp+oTSHc z7n_?1D+tU1K2Z5p16A+72aBvDoedV14n00JT%jMCLUZ3gT;4=W3_8X-F&3c02T00R1alVHrpcR; zrIg;+C9_Q;eiji63Ce2sny%-b?dNc1W)r%L>x-`WDN+4HAf7A;?z?7u#b`l~Ftga~r(oMBAJmry59@KD1plI>58Yx|ASXrnO 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