Chapitres Maths en ECG1
Chapitres Maths en ECG1
Corrigés : Variables aléatoires finies en ECG1
Résumé de cours Exercices Corrigés
Cours en ligne de Maths en ECG1
Corrigés – Variables aléatoires finies
Exercice 1 :
1) Pour réaliser on doit avoir
Ainsi
Par indépendance des événements, on a
Ainsi
2) a) Les deux derniers tirages sont Le premier tirage pouvant être Pile ou Face, on a bien
Ainsi
![Rendered by QuickLaTeX.com \mathbb{P} \left( X= 3 \right)](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-bb779ab2006aac1feeaeb1d951f3866e_l3.png)
![Rendered by QuickLaTeX.com = \mathbb{P} \left( P_1 P_2 F_3 \cup F_1 P_2 F_3 \right)](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-4ea8f518db5f1476877ce8849493609f_l3.png)
![Rendered by QuickLaTeX.com \text{les evenements etant disjoints}](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-cd1c09a1ceb64997d3fa12d9c8c3392f_l3.png)
![Rendered by QuickLaTeX.com = \mathbb{P} \left( P_1 P_2 F_3 \right) + \mathbb{P} \left( F_1 P_2 F_3 \right)](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-9eea0b58ea551da71d98f5e80ba229de_l3.png)
![Rendered by QuickLaTeX.com \text{par independance}](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-702bdb8d55cef3af41c629b9dfc2215a_l3.png)
![Rendered by QuickLaTeX.com =\mathbb{P} \left( P_1 \right) \times \mathbb{P} \left( P_2 \right) \times \mathbb{P} \left( F_3 \right)](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-2a11fe4b895eb2a0e7b5a9d157c10e6b_l3.png)
![Rendered by QuickLaTeX.com + \mathbb{P} \left( F_1 \right) \times \mathbb{P} \left( P_2 \right) \times \mathbb{P} \left( F_3 \right)](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-d66a6f87d7b0c17679ab99560acd1385_l3.png)
![Rendered by QuickLaTeX.com = \dfrac18 + \dfrac18](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-077a6a563a929342803336626c909ed8_l3.png)
![Rendered by QuickLaTeX.com = \dfrac14](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-76a813aa207a4b17db87f2b0078e1012_l3.png)
b) Les deux derniers lancers sont
On s’intéresse aux
premiers lancers. Deux possibilités :
![Rendered by QuickLaTeX.com P_{k - 1} F_k.](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-dcb35765226aaf3b4a391ab1bbd9a136_l3.png)
![Rendered by QuickLaTeX.com k - 2](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-f91a26b3f7bd650c49d6f903d249619e_l3.png)
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![Rendered by QuickLaTeX.com \star](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-ce5ce4f1ed0448af8dbee67daac86254_l3.png)
![Rendered by QuickLaTeX.com k - 2](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-f91a26b3f7bd650c49d6f903d249619e_l3.png)
![Rendered by QuickLaTeX.com \star](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-ce5ce4f1ed0448af8dbee67daac86254_l3.png)
![Rendered by QuickLaTeX.com k - 2](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-f91a26b3f7bd650c49d6f903d249619e_l3.png)
![Rendered by QuickLaTeX.com k- 2 \ge j \ge 1](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-950237a1a12cb451538964599ab4a381_l3.png)
![Rendered by QuickLaTeX.com j - 1](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-3cfdd5975bf300567ebdafad220816e4_l3.png)
![Rendered by QuickLaTeX.com j + 1, j + 2, \cdots , k- 2](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-c3181cefdb39ab99ec22f89182026bc8_l3.png)
En effet, si l’on avait un Face au cours de ces lancers. En notant
le rang d’apparition du premier Face au cours de ces lancers, on aurait un lancer du type
![Rendered by QuickLaTeX.com l](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-502276c66966e5a861539c7de60c26c0_l3.png)
Ainsi le rang d’apparition du premier groupement Pile puis Face serait en position
![Rendered by QuickLaTeX.com l < k,](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-a76b484907384edc02dfc193a1e28cfd_l3.png)
![Rendered by QuickLaTeX.com \left( X = k \right).](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-35320e1452dd9351308105c2ae977c3d_l3.png)
On a bien
![Rendered by QuickLaTeX.com \left( X= k \right)](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-9214b87760f24ba83cab6dd905a63cd2_l3.png)
![Rendered by QuickLaTeX.com = \left( F_1 F_2 \cdots F_{k - 2} P_{k - 1} F_k \right)](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-cbde1a5d181a3e82b0958fd00a3f55c2_l3.png)
![Rendered by QuickLaTeX.com \cup (\displaystyle\bigcup_{j=1}^{k - 2} \left( F_1 \cdots F_{j - 1} P_j P_{j + 1} \cdots P_{k - 1} F_k \right) ).](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-12fb3a0166c2b33e2094cdb01bac1c4d_l3.png)
Notons que ces
![Rendered by QuickLaTeX.com k - 1](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-ee62354264e7bacb9e0379203651c48d_l3.png)
c) A la question précédente, nous avons écrit
comme réunion de
événements deux à deux disjoints. Remarquons que, par indépendance des lancers, chacun de ces événements est de probabilité égale à
Les événements étant disjoints, on a:
![Rendered by QuickLaTeX.com \left( X = k \right)](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-787f5816684d3b7694c803685fac8664_l3.png)
![Rendered by QuickLaTeX.com k - 1](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-ee62354264e7bacb9e0379203651c48d_l3.png)
![Rendered by QuickLaTeX.com \left( \dfrac12 \right)^k.](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-e31e72f74f17b6700c4c785ca9c15317_l3.png)
![Rendered by QuickLaTeX.com \mathbb{P} \left( X = k \right)](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-31aae8397805b59da9e29ee3de5ec640_l3.png)
![Rendered by QuickLaTeX.com = \left( k - 1 \right) \left( \dfrac12 \right)^k.](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-008161cede9231e305f79394862783bf_l3.png)
d) Remarquons que
Ainsi
Si
se réalise. Comme le premier lancer est Pile,
se réalise. Ainsi,
se réalise car le premier groupement Pile puis Face est en position
et
Réciproquement, supposons que
se réalise. Montrons qu’il n’y a pas de Face lors dans lancers de rang
S’il y en a un, on note par
le rang d’apparition du premier Face. Ainsi, on a
![Rendered by QuickLaTeX.com + \mathbb{P} \left( X = k - 1 \right) \times \mathbb{P} \left( F_1 \right)](data:image/svg+xml,%3Csvg%20xmlns=%22http://www.w3.org/2000/svg%22%20viewBox=%220%200%20189%2018%22%3E%3C/svg%3E)
.
![Rendered by QuickLaTeX.com X \left( \Omega \right) = \mathbb{N} \backslash \left\{ 1 \right\}.](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-70f0a86a9aaaa9ec3b4081f0725e7220_l3.png)
.
3) a) On suppose que le premier lancer est Pile.
![Rendered by QuickLaTeX.com \bullet](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-2b6e225d778ccd32cb2bd9cc4eaead9a_l3.png)
![Rendered by QuickLaTeX.com P_2 P_3 \cdots P_{k- 1} F_k](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-81007a2aa98cf11d43298da014279961_l3.png)
![Rendered by QuickLaTeX.com P_1 P_2 \cdots P_{k - 1} F_k](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-17dfc151e9c8dacaa8e82f07e3936a84_l3.png)
![Rendered by QuickLaTeX.com \left( X = k \right)](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-787f5816684d3b7694c803685fac8664_l3.png)
![Rendered by QuickLaTeX.com k-1](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-1a4c925a5ad5141d0726a09015ceadcd_l3.png)
![Rendered by QuickLaTeX.com k.](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-4a22b46fd6f4018a6b70bc870f75be7b_l3.png)
![Rendered by QuickLaTeX.com \bullet](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-2b6e225d778ccd32cb2bd9cc4eaead9a_l3.png)
![Rendered by QuickLaTeX.com \left( X = k \right)](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-787f5816684d3b7694c803685fac8664_l3.png)
![Rendered by QuickLaTeX.com 2, 3 , \cdots, k - 2.](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-1166a38f9f9ac92e27fdb951a54e6955_l3.png)
![Rendered by QuickLaTeX.com j < k](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-970e5a9b568d6d97e37de77729ce0eef_l3.png)
![Rendered by QuickLaTeX.com P_1 P_2 \cdots P_{j - 1} F_j.](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-7fa8e81e02e44c096cfe2866e98c8da7_l3.png)
Ici, on remarque que le rang d’apparition du premier groupement Pile puis Face est au rang
ce qui est impossible car l’on travaille avec l’événement ![Rendered by QuickLaTeX.com \left( X = k \right).](data:image/svg+xml,%3Csvg%20xmlns=%22http://www.w3.org/2000/svg%22%20viewBox=%220%200%2069%2018%22%3E%3C/svg%3E)
![Rendered by QuickLaTeX.com j,](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-51223dde5149ec01a8a5c4b09e386482_l3.png)
![Rendered by QuickLaTeX.com \left( X = k \right).](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-35320e1452dd9351308105c2ae977c3d_l3.png)
b) Le famille
est un système complet d’événements. la formule des probabilités totales assure que
![Rendered by QuickLaTeX.com \left\{ P_1 , F_1 \right\}](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-74c4666e473e0443c82e2657b29b33f4_l3.png)
![Rendered by QuickLaTeX.com \mathbb{P} \left( X = k \right)](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-31aae8397805b59da9e29ee3de5ec640_l3.png)
![Rendered by QuickLaTeX.com = \mathbb{P}_{P_1} \left( X = k \right) \times \mathbb{P} \left( P_1 \right)](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-59239a9d313b841001a78af8bb99328a_l3.png)
![Rendered by QuickLaTeX.com + \mathbb{P}_{F_1} \left( X = k \right) \times \mathbb{P} \left( F_1 \right).](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-b18580bea46cdc44bb2f34bbec17fa97_l3.png)
Or, à la question précédente, on a montré que
remarquons aussi que lévénement
sachant
est équivalent à
ainsi
![Rendered by QuickLaTeX.com \mathbb{P}_{P_1} \left( X = k \right)](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-478d7ab29fa3665348b8cde972c60aae_l3.png)
![Rendered by QuickLaTeX.com = \mathbb{P} \left( P_2 P_3 \cdots P_{k - 1} F_k \right)](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-ff64bcb6ea27bcd73a40c0fc9bd20624_l3.png)
![Rendered by QuickLaTeX.com = \dfrac{1}{2^{k - 1}}.](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-d1cd4e7156af29cde786c7c7dcd5f392_l3.png)
![Rendered by QuickLaTeX.com \left( X = k \right)](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-787f5816684d3b7694c803685fac8664_l3.png)
![Rendered by QuickLaTeX.com F_1](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-f5288dee1beba163218d424e2d772e1d_l3.png)
![Rendered by QuickLaTeX.com \left( X = k - 1 \right),](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-98d3c65abffaa5c2b060ea15ba06898e_l3.png)
![Rendered by QuickLaTeX.com \mathbb{P} \left( X = k \right)](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-31aae8397805b59da9e29ee3de5ec640_l3.png)
![Rendered by QuickLaTeX.com = \dfrac{1}{2^{k - 1}} \times \mathbb{P} \left( P_1 \right)](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-d0197dd5300472725924807f186e3d8d_l3.png)
![Rendered by QuickLaTeX.com + \mathbb{P} \left( X = k - 1 \right) \times \mathbb{P} \left( F_1 \right)](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-0d03e6806bdad8cc69de31b48e90ef25_l3.png)
![Rendered by QuickLaTeX.com = \dfrac{1}{2^k} + \dfrac12 \mathbb{P} \left( X = k - 1 \right)](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-5c99212f727cdb232d95039f3308a96a_l3.png)
c) En multipliant la relation obtenue à la question précédente, on a:
![Rendered by QuickLaTeX.com = 2^{k - 1} \mathbb{P} \left( X = k - 1 \right) + 1,](data:image/svg+xml,%3Csvg%20xmlns=%22http://www.w3.org/2000/svg%22%20viewBox=%220%200%20196%2019%22%3E%3C/svg%3E)
soit
d’où
et
![Rendered by QuickLaTeX.com 2^k \mathbb{P} \left( X = k \right)](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-a2bb838fdaaae9f21ac8f459dcdc1d39_l3.png)
![Rendered by QuickLaTeX.com = 2^{k - 1} \mathbb{P} \left( X = k - 1 \right) + 1,](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-afcaeccc866054fcc9c53d5d5c917284_l3.png)
soit
d’où
![Rendered by QuickLaTeX.com u_k = k -1](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-706b9e242ca0811fea0f1efcd6fa129c_l3.png)
![Rendered by QuickLaTeX.com \mathbb{P} \left( X = k \right)](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-31aae8397805b59da9e29ee3de5ec640_l3.png)
![Rendered by QuickLaTeX.com = \dfrac{k - 1}{2^k}.](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-94b8e710195d6b7c02f8db64ffa48029_l3.png)
d) En multipliant la relation obtenue à la question précédente, on a:
![Rendered by QuickLaTeX.com = 2^{k - 1} \mathbb{P} \left( X = k - 1 \right) + 1,](data:image/svg+xml,%3Csvg%20xmlns=%22http://www.w3.org/2000/svg%22%20viewBox=%220%200%20196%2019%22%3E%3C/svg%3E)
soit
La suite
est arithmétique de raison
Pour tout
on a
d’où
et
![Rendered by QuickLaTeX.com 2^k \mathbb{P} \left( X = k \right)](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-a2bb838fdaaae9f21ac8f459dcdc1d39_l3.png)
![Rendered by QuickLaTeX.com = 2^{k - 1} \mathbb{P} \left( X = k - 1 \right) + 1,](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-afcaeccc866054fcc9c53d5d5c917284_l3.png)
soit
La suite
![Rendered by QuickLaTeX.com \left( u_k \right)_{k \ge 2}](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-e5bad8ab65331b6d41951f3512e38a96_l3.png)
![Rendered by QuickLaTeX.com 1.](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-cef179a64f3446ea0212dc931dea6fc8_l3.png)
![Rendered by QuickLaTeX.com k \ge 2,](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-1e4ccce9508b97e7e26ec8f62cd4ffd0_l3.png)
d’où
![Rendered by QuickLaTeX.com u_k = k -1](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-706b9e242ca0811fea0f1efcd6fa129c_l3.png)
![Rendered by QuickLaTeX.com \mathbb{P} \left( X = k \right)](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-31aae8397805b59da9e29ee3de5ec640_l3.png)
![Rendered by QuickLaTeX.com = \dfrac{k - 1}{2^k}.](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-94b8e710195d6b7c02f8db64ffa48029_l3.png)
4) Pour montrer que
admet une espérance, montrons que la série
converge absolument.
![Rendered by QuickLaTeX.com = \displaystyle\sum_{k=2}^{+ \infty} k \times \dfrac{k - 1}{2^k}](data:image/svg+xml,%3Csvg%20xmlns=%22http://www.w3.org/2000/svg%22%20viewBox=%220%200%20122%2053%22%3E%3C/svg%3E)
![Rendered by QuickLaTeX.com = \left( \dfrac12 \right)^2 \displaystyle\sum_{k=2}^{+ \infty} k \left( k - 1 \right) \left( \dfrac12 \right)^{k - 2}](data:image/svg+xml,%3Csvg%20xmlns=%22http://www.w3.org/2000/svg%22%20viewBox=%220%200%20231%2053%22%3E%3C/svg%3E)
![Rendered by QuickLaTeX.com = \dfrac14 \times \dfrac{2}{\left( 1 - \dfrac12 \right)^3}](data:image/svg+xml,%3Csvg%20xmlns=%22http://www.w3.org/2000/svg%22%20viewBox=%220%200%20133%2068%22%3E%3C/svg%3E)
.
![Rendered by QuickLaTeX.com X](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-d4ee28752517d6062a3ca0314890342d_l3.png)
![Rendered by QuickLaTeX.com \displaystyle\sum_{k \ge 2} k \mathbb{P} \left( X = k \right)](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-f25df21b0d7b05ab1dc74d52ea6744d3_l3.png)
Par croissance comparée, on a
![Rendered by QuickLaTeX.com \lim_{k \to + \infty} k^2 \times \dfrac{k - 1}{2^k} = 0.](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-3c2f57ccacc579d329215f5289806f99_l3.png)
Ainsi
Comme la série de terme général
converge (Riemann), la série
converge et
admet une espérance et
![Rendered by QuickLaTeX.com \dfrac{k -1}{2^{k - 1}} \underset{k \to + \infty}{=} o \left( \dfrac{1}{k^2} \right).](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-88cb380fb9bc697fff7166227eb336a2_l3.png)
![Rendered by QuickLaTeX.com \displaystyle\sum_{k \ge 2} \dfrac{1}{k^2}](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-dc821e4a1f2ce58b121cc94cc5d03309_l3.png)
![Rendered by QuickLaTeX.com \displaystyle\sum_{k \ge 2} k \mathbb{P} \left( X = k \right)](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-f25df21b0d7b05ab1dc74d52ea6744d3_l3.png)
![Rendered by QuickLaTeX.com X](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-d4ee28752517d6062a3ca0314890342d_l3.png)
![Rendered by QuickLaTeX.com \mathbb{E} \left( X \right)](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-bc47634680efc1ee69ff098a0e762687_l3.png)
![Rendered by QuickLaTeX.com = \displaystyle\sum_{k=2}^{+ \infty} k \times \dfrac{k - 1}{2^k}](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-11f99f03f422a6ffec2b41028119411a_l3.png)
![Rendered by QuickLaTeX.com = \left( \dfrac12 \right)^2 \displaystyle\sum_{k=2}^{+ \infty} k \left( k - 1 \right) \left( \dfrac12 \right)^{k - 2}](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-30da4edb21167ff71ee065baf92cfaf2_l3.png)
![Rendered by QuickLaTeX.com = \dfrac14 \times \dfrac{2}{\left( 1 - \dfrac12 \right)^3}](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-b2afc3c62245bd4e4d6d7f6c909d81c9_l3.png)
![Rendered by QuickLaTeX.com = 4](https://groupe-reussite.fr/ressources/wp-content/ql-cache/quicklatex.com-c061a2976dc53b0f43666b471984211f_l3.png)
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