File: /home/mmtprep/public_html/mathzen.mmtprep.com/assets/2F12-1-Zv1kMCsT.js.map
{"version":3,"file":"2F12-1-Zv1kMCsT.js","sources":["../../src/exercices/2e/2F12-1.js"],"sourcesContent":["import { choice, combinaisonListes } from '../../lib/outils/arrayOutils'\nimport { miseEnEvidence } from '../../lib/outils/embellissements'\nimport { extraireRacineCarree } from '../../lib/outils/calculs'\nimport { texFractionReduite } from '../../lib/outils/deprecatedFractions.js'\nimport { ecritureAlgebrique } from '../../lib/outils/ecritures'\nimport { sp } from '../../lib/outils/outilString.js'\nimport { texNombre } from '../../lib/outils/texNombre'\nimport Exercice from '../deprecatedExercice.js'\nimport {\n listeQuestionsToContenu, randint\n} from '../../modules/outils.js'\nimport { fraction } from '../../modules/fractions.js'\nexport const titre = 'Algebraically solve an equation f(x)=k with a reference function'\nexport const dateDePublication = '07/01/2022'\n/**\n*\n*\n* @author Gilles Mora // suppression des calcul des texNombre et simplification des racines carrées de fration par Jean-Claude Lhote\n*\n*/\nexport const uuid = 'de0d1'\nexport const ref = '2F12-1'\nexport default function EquationsFonctionsRef () {\n Exercice.call(this) // Héritage de la classe Exercice()\n this.sup = 1\n this.sup2 = 1\n this.consigne = ''\n this.correctionDetailleeDisponible = true\n this.correctionDetaillee = false\n this.spacing = 1\n this.nbQuestions = 2\n this.nbQuestionsModifiable = true\n this.nouvelleVersion = function () {\n this.autoCorrection = []\n this.sup = parseInt(this.sup)\n this.listeQuestions = [] // Liste de questions\n this.listeCorrections = [] // Liste de questions corrigées\n let typesDeQuestionsDisponibles\n switch (this.sup) {\n case 1:\n typesDeQuestionsDisponibles = ['x^2=k']\n break\n case 2:\n typesDeQuestionsDisponibles = ['sqrt(x)=k']\n break\n case 3:\n typesDeQuestionsDisponibles = ['1/x=k']\n break\n case 4:\n typesDeQuestionsDisponibles = ['x^3=k']\n break\n case 5:\n typesDeQuestionsDisponibles = ['x^2=k', 'sqrt(x)=k', '1/x=k', 'x^3=k']\n break\n //\n }\n function ecritureParentheseSiNegatif (a, maximumFractionDigits = 15) {\n const result = Intl.NumberFormat('fr-FR', { maximumFractionDigits }).format(a).replace(',', '{,}')\n return a < 0 ? `(${result})` : result\n }\n const listeTypeDeQuestions = combinaisonListes(typesDeQuestionsDisponibles, this.nbQuestions)\n let sousChoix\n if (parseInt(this.sup2) === 1) {\n sousChoix = combinaisonListes([0], this.nbQuestions) // pour choisir aléatoirement des questions dans chaque catégorie\n } else if (parseInt(this.sup2) === 2) {\n sousChoix = combinaisonListes([1, 2, 3], this.nbQuestions)\n } else {\n sousChoix = combinaisonListes([0, 1, 2, 3], this.nbQuestions)\n }\n for (let i = 0, texte, texteCorr, x, y, a, b, c, k, k1, f1, enonce, correction, cpt = 0; i < this.nbQuestions && cpt < 50;) {\n // we only choose numbers between 1 and 20\n x = randint(-9, 9, [0, 1, -1])\n y = randint(-9, 9, [x, 0])\n\n switch (listeTypeDeQuestions[i]) {\n case 'x^2=k':\n switch (sousChoix[i]) { //\n case 0:\n a = randint(0, 15) ** 2\n k = choice([randint(-20, 50, [1, 4, 9, 16, 25, 36, 49]), randint(-20, 50, [1, 4, 9, 16, 25, 36, 49]), randint(-20, 50, [1, 4, 9, 16, 25, 36, 49]), a])\n enonce = `Solve in $\\\\mathbb{R}$:<br>${sp(50)} $x^2=${k}$`\n correction = ''\n if (this.correctionDetaillee) {\n correction += 'The equation $x^2=k$ admits:<br>$\\\\bullet~$ two solutions when $k>0$: $-\\\\sqrt{k}$ and $\\\\sqrt{k}$ ;<br>$\\\\bullet~$ a unique solution equal to $0$ when $k=0$; <br>$\\\\bullet~$ no solution when $k<0$.<br><br>'\n }\n if (k > 0) {\n correction += `The equation is of the form $x^2=k$ with $k=${k}$. As $${k}>0$ then the equation admits two solutions: $-\\\\sqrt{${k}}$ and $\\\\sqrt{${k}}$.<br>`\n if (extraireRacineCarree(k)[1] === k) {\n if (k === 1) {\n correction += `As $-\\\\sqrt{${k}}=-${Math.sqrt(k)}$ and $\\\\sqrt{${k}}=${Math.sqrt(k)}$ then the solutions of the equation can be written more simply: $-${Math.sqrt(k)}$ and $${Math.sqrt(k)}$.<br>Thus , $S=\\\\{-${Math.sqrt(k)},${sp(1)}${Math.sqrt(k)}\\\\}$. `\n } else {\n correction += `So, $S=\\\\{-\\\\sqrt{${k}},${sp(1)}\\\\sqrt{${k}}\\\\}$. `\n }\n } else {\n if (k === a) {\n correction += `As $-\\\\sqrt{${k}}=-${Math.sqrt(k)}$ and $\\\\sqrt{${k}}=${Math.sqrt(k)}$ then \n the solutions of the equation can be written more simply: $-${Math.sqrt(k)}$ and $${Math.sqrt(k)}$.<br>\n Thus , $S=\\\\{-${Math.sqrt(k)},${sp(1)}${Math.sqrt(k)}\\\\}$. `\n } else {\n correction += `As $-\\\\sqrt{${k}}=-${extraireRacineCarree(k)[0]}\\\\sqrt{${extraireRacineCarree(k)[1]}}$ and $\\\\sqrt{${k}}=${extraireRacineCarree(k)[0]}\\\\sqrt{${extraireRacineCarree(k)[1]}}$ then the solutions of the equation can be written more simply: $-${extraireRacineCarree(k)[0]}\\\\sqrt{${extraireRacineCarree(k)[1]}}$ and $${extraireRacineCarree(k)[0]}\\\\sqrt{${extraireRacineCarree(k)[1]}}$.<br>Thus, $S=\\\\{-${extraireRacineCarree(k)[0]}\\\\sqrt{${extraireRacineCarree(k)[1]}},${sp(1)}${extraireRacineCarree(k)[0]}\\\\sqrt{${extraireRacineCarree(k)[1]}}\\\\}$. `\n }\n }\n } else {\n if (k === 0) {\n correction += `The equation is of the form $x^2=k$ with $k=${k}$. As $k=${k}$ then The equation admits a unique solution: $0$.<br>Thus, $S=\\\\{0\\\\}$. `\n } else {\n correction += `The equation is of the form $x^2=k$ with $k=${k}$. Since $${k}<0$, then the equation admits no solution.<br>Thus, $S=\\\\emptyset$. `\n }\n }\n\n break\n case 1:// x^2+b=c\n b = randint(-15, 15, 0)\n c = randint(-15, 15, 0)\n k = c - b\n enonce = `Solve in $\\\\mathbb{R}$:<br>${sp(50)} $x^2${ecritureAlgebrique(b)}=${c}$`\n correction = 'We isolate $x^2$ on the left side to obtain an equation of the type $x^2=k$.<br>'\n if (b > 0) {\n correction += `$\\\\begin{aligned}x^2${ecritureAlgebrique(b)}&=${c}\\\\\\\\x^2${ecritureAlgebrique(b)}-${miseEnEvidence(b)}&=${c}-${miseEnEvidence(b)}\\\\\\\\x^2&=${c - b}\\\\end{aligned}$`\n } else {\n correction += `$\\\\begin{aligned}x^2${ecritureAlgebrique(b)}&=${c}\\\\\\\\x^2${ecritureAlgebrique(b)}+${miseEnEvidence(-b)}&=${c}+${miseEnEvidence(-b)}\\\\\\\\x^2&=${c - b}\\\\end{aligned}$`\n }\n if (k > 0) {\n if (k === 1 || k === 4 || k === 9 || k === 16 || k === 25) {\n correction += `<br>The equation is of the form $x^2=k$ with $k=${texNombre(k, 0)}$. As $${texNombre(k, 0)}>0$, the equation has two solutions: $-\\\\sqrt{${texNombre(k, 0)}}$ and $\\\\sqrt{${texNombre(k, 0)}}$.<br> As $-\\\\sqrt{${texNombre(k, 0)}}=-${extraireRacineCarree(k)[0]}$ and $\\\\sqrt{${k}}=${extraireRacineCarree(k)[0]}$ then the solutions of the equation can be written more simply: $-${extraireRacineCarree(k)[0]}$ and $${extraireRacineCarree(k)[0]}$.<br>Thus, $S=\\\\{-${extraireRacineCarree(k)[0]},${sp(1)}${extraireRacineCarree(k)[0]}\\\\} $. `\n } else {\n if (extraireRacineCarree(k)[1] !== k) {\n correction += `<br>The equation is of the form $x^2=k$ with $k=${texNombre(k, 0)}$. Since $${texNombre(k, 0)}>0$, the equation has two solutions: $-\\\\sqrt{${texNombre(k, 0)}}$ and $\\\\sqrt{${texNombre(k, 0)}}$. <br>As $-\\\\sqrt{${k}}=-${extraireRacineCarree(k)[0]}\\\\sqrt{${extraireRacineCarree(k)[1]}}$ and $\\\\sqrt{${k}}=${extraireRacineCarree(k)[0]}\\\\sqrt{${extraireRacineCarree(k)[1]}}$ then the solutions of the equation can be written more simply: $-${extraireRacineCarree(k)[0]}\\\\sqrt{${extraireRacineCarree(k)[1]}}$ and $${extraireRacineCarree(k)[0]}\\\\sqrt{${extraireRacineCarree(k)[1]}}$.<br>Thus, $S=\\\\{-${extraireRacineCarree(k)[0]}\\\\sqrt{${extraireRacineCarree(k)[1]}},${sp(1)}${extraireRacineCarree(k)[0]}\\\\sqrt{${extraireRacineCarree(k)[1]} }\\\\}$. `\n } else {\n correction += `<br>The equation is of the form $x^2=k$ with $k=${c - b}$. Since $${c - b}>0$, the equation has two solutions: $-\\\\sqrt{${c - b}}$ and $\\\\sqrt{${c - b}}$.<br>Thus, $S=\\\\{-\\\\sqrt{${c - b} },${sp(1)}\\\\sqrt{${c - b}}\\\\}$. `\n }\n }\n }\n if (k === 0) {\n correction += `<br>The equation is of the form $x^2=k$ with $k=${texNombre(k, 0)}$, then the equation has a solution: $0$.<br>Thus, $S=\\\\{0\\\\} $. `\n }\n if (k < 0) {\n correction += `<br>The equation is of the form $x^2=k$ with $k=${texNombre(c - b, 0)}$. Since $${texNombre(c - b, 0)}<0$, the equation has no solution.<br>Thus, $S=\\\\emptyset$. `\n }\n break\n case 2:// -x^2+b=c\n b = randint(-10, 10, 0)\n c = randint(-10, 10, 0)\n k = b - c\n enonce = `Solve in $\\\\mathbb{R}$:<br>${sp(50)} $-x^2${ecritureAlgebrique(b)}=${c}$`\n correction = 'We isolate $x^2$ on the left side to obtain an equation of the type $x^2=k$.<br>'\n if (b > 0) {\n correction += `$\\\\begin{aligned}-x^2+${b}&=${c}\\\\\\\\-x^2${ecritureAlgebrique(b)}-${miseEnEvidence(b)}&=${c}-${miseEnEvidence(b)}\\\\\\\\-x^2&=${c - b}\\\\\\\\x^2&=${b - c}\\\\ end{aligned}$`\n } else {\n correction += `$\\\\begin{aligned}-x^2${ecritureAlgebrique(b)}&=${c}\\\\\\\\-x^2${ecritureAlgebrique(b)}+${miseEnEvidence(-b)}&=${c}+${miseEnEvidence(-b)}\\\\\\\\-x^2&=${c - b}\\\\\\\\x^2&=${b - c}\\\\end{ aligned}$`\n }\n\n if (k > 0) {\n if (k === 1 || k === 4 || k === 9 || k === 16 || k === 25) {\n correction += `<br>The equation is of the form $x^2=k$ with $k=${texNombre(k, 0)}$. As $${texNombre(k, 0)}>0$, the equation has two solutions: $-\\\\sqrt{${texNombre(k, 0)}}$ and $\\\\sqrt{${texNombre(k, 0)}}$.<br> As $-\\\\sqrt{${texNombre(k, 0)}}=-${extraireRacineCarree(k)[0]}$ and $\\\\sqrt{${k}}=${extraireRacineCarree(k)[0]}$ then the solutions of the equation can be written more simply: $-${extraireRacineCarree(k)[0]}$ and $${extraireRacineCarree(k)[0]}$.<br>Thus, $S=\\\\{-${extraireRacineCarree(k)[0]},${sp(1)}${extraireRacineCarree(k)[0]}\\\\} $. `\n } else {\n if (extraireRacineCarree(k)[1] !== k) {\n correction += `<br>The equation is of the form $x^2=k$ with $k=${texNombre(k, 0)}$. As $${texNombre(k, 0)}>0$, the equation has two solutions: $-\\\\sqrt{${texNombre(k, 0)}}$ and $\\\\sqrt{${texNombre(k, 0)}}$.<br>As $-\\\\sqrt{${k}}=-${extraireRacineCarree(k)[0]}\\\\ sqrt{${extraireRacineCarree(k)[1]}}$ and $\\\\sqrt{${k}}=${extraireRacineCarree(k)[0]}\\\\sqrt{${extraireRacineCarree(k)[1]}}$ then the solutions of the equation can be written more simply: $-${extraireRacineCarree(k)[0]}\\\\sqrt{${extraireRacineCarree(k)[1]}}$ and $${extraireRacineCarree(k)[0]}\\\\ sqrt{${extraireRacineCarree(k)[1]}}$.<br>Thus, $S=\\\\{-${extraireRacineCarree(k)[0]}\\\\sqrt{${extraireRacineCarree(k)[1]}},${sp(1)}${extraireRacineCarree(k)[0]}\\\\sqrt{${extraireRacineCarree(k)[1]}}\\\\}$. `\n } else {\n correction += `<br>The equation is of the form $x^2=k$ with $k=${texNombre(k, 0)}$. Since $${texNombre(k, 0)}>0$, then the equation has two solutions: $-\\\\sqrt{${texNombre(k, 0)}}$ and $\\\\sqrt{${texNombre(k, 0)}}$.<br>Thus, $S=\\\\{-\\\\sqrt{ ${k}},${sp(1)}\\\\sqrt{${k}}\\\\}$. `\n }\n }\n }\n if (k === 0) {\n correction += `<br>The equation is of the form $x^2=k$ with $k=${texNombre(k, 0)}$, so the equation has a solution: $0$.<br>Thus, $S=\\\\{0\\\\} $. `\n }\n if (k < 0) {\n correction += `<br>The equation is of the form $x^2=k$ with $k=${texNombre(b - c)}$. Since $${texNombre(b - c)}<0$, the equation has no solution.<br> Thus, $S=\\\\emptyset$. `\n }\n break\n\n case 3:// ax^2+b=c\n a = randint(-10, 10, [-1, 0, 1])\n b = randint(-10, 10, 0)\n c = randint(-10, 10, 0)\n k = (c - b) / a\n f1 = fraction(c - b, a)\n enonce = `Solve in $\\\\mathbb{R}$:<br>${sp(50)} $${a}x^2${ecritureAlgebrique(b)}=${c}$`\n correction = 'We isolate $x^2$ on the left side to obtain an equation of the type $x^2=k$.<br>'\n if (b > 0) {\n correction += `$\\\\begin{aligned}${a}x^2${ecritureAlgebrique(b)}&=${c}\\\\\\\\${a}x^2${ecritureAlgebrique(b)}-${miseEnEvidence(b)}&=${c}-${miseEnEvidence(b)}\\\\\\\\${a}x^2&=${c - b}\\\\\\\\x^2&=${texFractionReduite(c - b, a)}\\\\end{aligned}$`\n } else {\n correction += `$\\\\begin{aligned}${a}x^2${ecritureAlgebrique(b)}&=${c}\\\\\\\\${a}x^2${ecritureAlgebrique(b)}+${miseEnEvidence(-b)}&=${c}+${miseEnEvidence(-b)}\\\\\\\\${a}x^2&=${c - b}\\\\\\\\x^2&=${texFractionReduite(c - b, a)}\\\\end{aligned}$`\n }\n if (k > 0) {\n if (c - b === a || c - b === 4 * a || c - b === 9 * a || c - b === 16 * a || c - b === 25 * a) {\n correction += `<br>The equation is of the form $x^2=k$ with $k=${texNombre(k, 0)}$. As $${texNombre(k, 0)}>0$, then the equation has two solutions: $-\\\\sqrt{${texNombre(k, 0)}}$ and $\\\\sqrt{${texNombre(k, 0)}}$.<br> As $-\\\\sqrt{${texNombre(k, 0)}}=-${extraireRacineCarree(k)[0]}$ and $\\\\sqrt{${k}}=${extraireRacineCarree(k)[0]}$ then the solutions of the equation can be written more simply: $-${extraireRacineCarree(k)[0]}$ and $${extraireRacineCarree(k)[0]}$.<br> Thus, $S=\\\\left\\\\{-${extraireRacineCarree(k)[0]}, ${sp(1)}${extraireRacineCarree(k)[0]}\\\\right\\\\}$. `\n } else {\n if (((c - b === 4) && a === 9) || ((c - b === 9) && a === 4) || ((c - b === 16) && a === 9) || ((c - b === 9) && a === 16)) {\n correction += `<br>The equation is of the form $x^2=k$ with $k=${texFractionReduite(c - b, a)}$. As $${texFractionReduite(c - b, a)}>0$, then the equation has two solutions: $-\\\\sqrt{${texFractionReduite(c - b, a)}}$ and $\\\\sqrt{${texFractionReduite(c - b, a)}}$.<br> As $-\\\\sqrt{${texFractionReduite(c - b, a)}}=-\\\\ dfrac{${extraireRacineCarree(c - b)[0]}}{${extraireRacineCarree(a)[0]}}$ and $\\\\sqrt{${texFractionReduite(c - b, a)}}=\\\\dfrac{${extraireRacineCarree(c - b)[0]}}{${extraireRacineCarree(a)[0]}}$ then the solutions of the equation can be written more simply: $-\\\\dfrac{${extraireRacineCarree(c - b)[0]}}{${extraireRacineCarree(a)[0]} }$ and $\\\\dfrac{${extraireRacineCarree(c - b)[0]}}{${extraireRacineCarree(a)[0]}}$.<br>Thus, $S=\\\\left\\\\{-\\\\dfrac{${extraireRacineCarree(c - b)[0]}}{${extraireRacineCarree(a)[0]}},${sp(1)}\\\\dfrac{${extraireRacineCarree(c - b)[0]}}{${extraireRacineCarree(a)[0]} }\\\\right\\\\}$`\n } else {\n correction += `<br>The equation is of the form $x^2=k$ with $k=${texFractionReduite(c - b, a)}$. Since $${texFractionReduite(c - b, a)}>0$, then the equation has two solutions: $-${f1.racineCarree().texFractionSimplifiee}$ and $${f1.racineCarree().texFractionSimplifiee}$. <br>Thus, $S=\\\\left\\\\{-${f1.racineCarree().texFractionSimplifiee},${sp(1)}${f1.racineCarree().texFractionSimplifiee}\\\\right\\\\}$`\n }\n }\n }\n\n if (c - b === 0) {\n correction += '<br>The equation is of the form $x^2=k$ with $k=0$. Then the equation has a solution: $0$.<br>Thus, $S=\\\\{0\\\\}$.'\n }\n if ((c - b) / a < 0) {\n correction += `<br>The equation is of the form $x^2=k$ with $k=${texFractionReduite(c - b, a)}$. Since $${texFractionReduite(c - b, a)}<0$, then the equation has no solution. <br>Thus, $S=\\\\emptyset$. `\n }\n\n break\n }\n break\n case 'sqrt(x)=k':\n switch (sousChoix[i]) {\n case 0:// sqrt(x)=k\n k = randint(-25, 25, 0)\n enonce = `Solve in $[0,${sp(1)}+\\\\infty[$:<br>${sp(50)} $\\\\sqrt{x}=${k}$`\n correction = ''\n if (this.correctionDetaillee) {\n correction += 'For any positive or zero real $x$, the equation $\\\\sqrt{x}=k$ admits:<br>$\\\\bullet~$ a solution if $k\\\\geqslant 0$: $k^2$ ;<br>$\\\\bullet~$ no solution if $k<0$.<br>'\n }\n if (k < 0) {\n correction += `The equation is of the form $\\\\sqrt{x}=k$. As $k=${k}$ and $${k}<0$ then the equation has no solution.<br>Thus, $S=\\\\emptyset$. `\n }\n if (k > 0 || k === 0) {\n correction += `$k=${k}$ and $${k}>0$ therefore the equation admits a solution: $${k}^2=${k ** 2}$.<br>Thus $S=\\\\{${k ** 2}\\\\}$. `\n }\n break\n\n case 1:// sqrt(x)+b=c\n b = randint(-10, 10, 0)\n c = randint(-10, 10)\n k = c - b\n enonce = `Solve in $[0,${sp(1)}+\\\\infty[$:<br>${sp(50)} $\\\\sqrt{x}${ecritureAlgebrique(b)}=${c}$`\n\n if (b > 0) {\n correction = `We isolate $\\\\sqrt{x}$ on the left side to obtain an equation of the type $\\\\sqrt{x}=k$.<br>$\\\\begin{aligned}\\\\sqrt{x}${ecritureAlgebrique(b)}&= ${c}\\\\\\\\\\\\sqrt{x}${ecritureAlgebrique(b)}-${miseEnEvidence(b)}&=${c}-${miseEnEvidence(b)}\\\\\\\\\\\\sqrt{x}&=${c - b}\\\\end{aligned}$<br>`\n } else {\n correction = `We isolate $\\\\sqrt{x}$ on the left side to obtain an equation of the type $\\\\sqrt{x}=k$.<br>$\\\\begin{aligned}\\\\sqrt{x}${ecritureAlgebrique(b)}&= ${c}\\\\\\\\\\\\sqrt{x}${ecritureAlgebrique(b)}+${miseEnEvidence(-b)}&=${c}+${miseEnEvidence(-b)}\\\\\\\\\\\\sqrt{x}&=${c - b}\\\\end{aligned}$<br>`\n }\n if (c - b < 0) {\n correction += `The equation is of the form $\\\\sqrt{x}=k$ with $k=${k}$. As $${k}<0$ then the equation has no solution. <br>So, $S=\\\\emptyset$.<br>`\n }\n if (c - b > 0 || c - b === 0) {\n correction += `The equation is of the form $\\\\sqrt{x}=k$ with $k=${c - b}$. As $${c - b}\\\\geqslant 0$ then the equation admits a solution: $${k}^2=${k ** 2}$.<br>Thus $S=\\\\{${k ** 2}\\\\}$. `\n }\n\n break\n case 2:// -sqrt(x)+b=c\n b = randint(-10, 10, 0)\n c = randint(-10, 10)\n k = b - c\n enonce = `Solve in $[0,${sp(1)}+\\\\infty[$:<br>${sp(50)} $-\\\\sqrt{x}${ecritureAlgebrique(b)}=${c}$`\n if (b > 0) {\n correction = `We isolate $\\\\sqrt{x}$ on the left side to obtain an equation of the type $\\\\sqrt{x}=k$.<br>$\\\\begin{aligned}-\\\\sqrt{x}${ecritureAlgebrique(b)}& =${c}\\\\\\\\-\\\\sqrt{x}${ecritureAlgebrique(b)}-${miseEnEvidence(b)}&=${c}-${miseEnEvidence(b)}\\\\\\\\-\\\\sqrt{x}&=${c - b}\\\\\\\\\\\\sqrt{x}&=${b - c}\\\\end {aligned}$<br>`\n } else {\n correction = `We isolate $\\\\sqrt{x}$ on the left side to obtain an equation of the type $\\\\sqrt{x}=k$.<br>$\\\\begin{aligned}-\\\\sqrt{x}${ecritureAlgebrique(b)}& =${c}\\\\\\\\-\\\\sqrt{x}${ecritureAlgebrique(b)}+${miseEnEvidence(-b)}&=${c}+${miseEnEvidence(-b)}\\\\\\\\-\\\\sqrt{x}&=${c - b}\\\\\\\\\\\\sqrt{x}&=${b - c}\\\\end {aligned}$<br>`\n }\n if (k < 0) {\n correction += `The equation is of the form $\\\\sqrt{x}=k$ with $k=${k}$. As $${k}<0$ then the equation has no solution. <br>So, $S=\\\\emptyset$.<br>`\n }\n if (k > 0 || k === 0) {\n correction += `The equation is of the form $\\\\sqrt{x}=k$ with $k=${b - c}$. As $${b - c}\\\\geqslant0$ then the equation admits a solution: $${k}^2=${k ** 2}$.<br>Thus $S=\\\\{${k ** 2}\\\\}$. `\n }\n\n break\n case 3:// a*sqrt(x)+b=c\n a = randint(-10, 10, [0, -1, 1])\n b = randint(-10, 10, 0)\n c = randint(-10, 10)\n k = (c - b) / a\n enonce = `Solve in $[0,${sp(1)}+\\\\infty[$:<br>${sp(50)} $${a}\\\\sqrt{x}${ecritureAlgebrique(b)}=${c}$`\n if (b > 0) {\n correction = `We isolate $\\\\sqrt{x}$ in the left side to obtain an equation of the type $\\\\sqrt{x}=k$.<br>$\\\\begin{aligned}${a}\\\\sqrt{x}${ecritureAlgebrique(b)}& =${c}\\\\\\\\${a}\\\\sqrt{x}${ecritureAlgebrique(b)}-${miseEnEvidence(b)}&=${c}-${miseEnEvidence(b)}\\\\\\\\${a}\\\\sqrt{x}&=${c - b}\\\\\\\\\\\\sqrt{x}&=${texFractionReduite(c - b, a)}\\\\end {aligned}$<br>`\n } else {\n correction = `We isolate $\\\\sqrt{x}$ in the left side to obtain an equation of the type $\\\\sqrt{x}=k$.<br>$\\\\begin{aligned}${a}\\\\sqrt{x}${ecritureAlgebrique(b)}& =${c}\\\\\\\\${a}\\\\sqrt{x}${ecritureAlgebrique(b)}+${miseEnEvidence(-b)}&=${c}+${miseEnEvidence(-b)}\\\\\\\\${a}\\\\sqrt{x}&=${c - b}\\\\\\\\\\\\sqrt{x}&=${texFractionReduite(c - b, a)}\\\\end {aligned}$<br>`\n }\n if (k < 0) {\n correction += `The equation is of the form $\\\\sqrt{x}=k$ with $k=${texFractionReduite(c - b, a)}$. As $${texFractionReduite(c - b, a)}<0$ then the equation has no solution. <br>So, $S=\\\\emptyset$.<br>`\n }\n if (k > 0 || k === 0) {\n correction += `The equation is of the form $\\\\sqrt{x}=k$ with $k=${texFractionReduite(c - b, a)}$. As $${texFractionReduite(c - b, a)}\\\\geqslant0$ then the equation admits a solution: $\\\\left(${texFractionReduite(c - b, a)}\\\\right)^2=${texFractionReduite((c - b) ** 2, a ** 2)}$.<br>Thus $S=\\\\left\\\\{${texFractionReduite((c - b) ** 2, a ** 2)}\\\\right\\\\ }$. `\n }\n\n break\n }\n break\n case '1/x=k':\n\n switch (sousChoix[i]) { // sousChoix[i] = randint(0, 5)\n case 0:\n k = randint(-10, 10)\n\n enonce = `Solve in $\\\\mathbb{R}^*$:<br>${sp(50)} $\\\\dfrac{1}{x}=${k}$`\n correction = ''\n if (this.correctionDetaillee) {\n correction += 'The equation $\\\\dfrac{1}{x}=k$ admits:<br>$\\\\bullet~$ if $k\\\\neq 0$, the equation has a unique solution: $\\\\dfrac{1 }{k}$.<br>$\\\\bullet~$ if $k= 0$, the equation has no solution.<br>'\n }\n correction += ''\n if (k === 0) {\n correction += `The equation is of the form $\\\\dfrac{1}{x}=k$ with $k=${k}$. Since $k=${k}$, then the equation has no solution.<br>Thus, $S=\\\\emptyset$. `\n }\n if (k !== 0) {\n correction += `The equation is of the form $\\\\dfrac{1}{x}=k$ with $k=${k}$. As $${k}\\\\neq 0$ then the equation admits a solution:$${texFractionReduite(1, k)}$.<br>Thus $S=\\\\left\\\\{${texFractionReduite(1, k)}\\\\right\\\\}$. `\n }\n\n break\n\n case 1:\n b = randint(-10, 10, 0)\n c = randint(-10, 10)\n k = c - b\n enonce = `Solve in $\\\\mathbb{R}^*$:<br>${sp(50)} $\\\\dfrac{1}{x}${ecritureAlgebrique(b)}=${c}$`\n correction = ''\n if (b > 0) {\n correction += `We isolate $\\\\dfrac{1}{x}$ on the left side to obtain an equation of the type $\\\\dfrac{1}{x}=k$.<br>$\\\\begin{aligned}\\\\ dfrac{1}{x}${ecritureAlgebrique(b)}&=${c}\\\\\\\\\\\\dfrac{1}{x}${ecritureAlgebrique(b)}-${miseEnEvidence(b)}&=${c}-${miseEnEvidence(b)}\\\\\\\\\\\\dfrac{1}{x}&=${c - b}\\\\end{ aligned}$<br>`\n } else {\n correction += `We isolate $\\\\dfrac{1}{x}$ on the left side to obtain an equation of the type $\\\\dfrac{1}{x}=k$.<br>$\\\\begin{aligned}\\\\ dfrac{1}{x}${ecritureAlgebrique(b)}&=${c}\\\\\\\\\\\\dfrac{1}{x}${ecritureAlgebrique(b)}+${miseEnEvidence(-b)}&=${c}+${miseEnEvidence(-b)}\\\\\\\\\\\\dfrac{1}{x}&=${c - b}\\\\end{ aligned}$<br>`\n }\n if (k === 0) {\n correction += `The equation is of the form $\\\\dfrac{1}{x}=k$ with $k=${k}$. So the equation has no solution.<br>Thus, $S=\\\\emptyset$. `\n }\n if (k !== 0) {\n correction += `$k=${k}$ and $${k}\\\\neq 0$, so the equation is of the form $\\\\dfrac{1}{x}=k$ with $k=${k}$. So the equation admits a solution:$${texFractionReduite(1, k)}$.<br>Thus $S=\\\\left\\\\{${texFractionReduite(1, k)}\\\\right\\\\}$. `\n }\n break\n case 2:\n a = randint(-10, 10, 0)\n b = randint(-10, 10, 0)\n k = b / a\n enonce = `Solve in $\\\\mathbb{R}^*$:<br>${sp(50)} $\\\\dfrac{${a}}{x}=${b}$`\n correction = ''\n correction += `We isolate $\\\\dfrac{1}{x}$ on the left side to obtain an equation of the type $\\\\dfrac{1}{x}=k$.<br>$\\\\begin{aligned}\\\\ dfrac{${a}}{x}&=${b}\\\\\\\\\\\\dfrac{1}{x}&=${texFractionReduite(b, a)}${sp(20)}\\\\text{By dividing by } ${a} \\\\text{ in each member}\\\\\\\\\\\\end {aligned}$<br>`\n if (k === 0) {\n correction += `The equation is of the form $\\\\dfrac{1}{x}=k$ with $k=${k}$. So the equation has no solution.<br>Thus, $S=\\\\emptyset$. `\n }\n if (k !== 0) {\n if (k % 1 === 0) {\n correction += `The equation is of the form $\\\\dfrac{1}{x}=k$ with $k=${texFractionReduite(b, a)}$. So the equation admits a solution:$\\\\dfrac{1}{${texFractionReduite(b, a)}}$.<br>Thus $S=\\\\left\\\\{${texFractionReduite(a, b)}\\\\right\\\\}$. `\n } else {\n correction += `The equation is of the form $\\\\dfrac{1}{x}=k$ with $k=${texFractionReduite(b, a)}$. So the equation admits a solution:$\\\\dfrac{1}{${texFractionReduite(b, a)}}=${texFractionReduite(a, b)}$.<br>Thus $S=\\\\left\\\\{${texFractionReduite(a, b)}\\\\right\\\\}$. `\n }\n }\n break\n case 3:\n a = randint(-10, 10, 0)\n b = randint(-10, 10, 0)\n c = randint(-10, 10, 0)\n k = (c - b) / a\n enonce = `Solve in $\\\\mathbb{R}^*$:<br>${sp(50)} $\\\\dfrac{${a}}{x}${ecritureAlgebrique(b)}=${c}$`\n correction = ''\n if (b > 0) {\n correction += `We isolate $\\\\dfrac{1}{x}$ on the left side to obtain an equation of the type $\\\\dfrac{1}{x}=k$.<br>$\\\\begin{aligned}\\\\ dfrac{${a}}{x}${ecritureAlgebrique(b)}&=${c}\\\\\\\\\\\\dfrac{${a}}{x}${ecritureAlgebrique(b)}-${miseEnEvidence(b)}&=${c}-${miseEnEvidence(b)}\\\\\\\\\\\\dfrac{${a}}{x}&=${c - b}\\\\\\\\\\\\dfrac{1}{x}&=${texFractionReduite(c - b, a)}${sp(20)}\\\\text{By dividing by } ${a} \\\\text{ in each member}\\\\end{aligned}$<br>`\n } else {\n correction += `We isolate $\\\\dfrac{1}{x}$ on the left side to obtain an equation of the type $\\\\dfrac{1}{x}=k$.<br>$\\\\begin{aligned}\\\\ dfrac{${a}}{x}${ecritureAlgebrique(b)}&=${c}\\\\\\\\\\\\dfrac{${a}}{x}${ecritureAlgebrique(b)}+${miseEnEvidence(-b)}&=${c}+${miseEnEvidence(-b)}\\\\\\\\\\\\dfrac{${a}}{x}&=${c - b}\\\\\\\\\\\\dfrac{1}{x}&=${texFractionReduite(c - b, a)}${sp(20)}\\\\text{By dividing by } ${a} \\\\text{ in each member}\\\\\\\\\\\\end{aligned}$<br>`\n }\n if (k === 0) {\n correction += `The equation is of the form $\\\\dfrac{1}{x}=k$ with $k=${texFractionReduite(c - b, a)}$. So the equation has no solution.<br>Thus, $S=\\\\emptyset$. `\n }\n if (k !== 0) {\n if (k % 1 === 0) {\n correction += `The equation is of the form $\\\\dfrac{1}{x}=k$ with $k=${texFractionReduite(c - b, a)}$. So the equation admits a solution:$\\\\dfrac{1}{${texFractionReduite(c - b, a)}}$.<br>Thus $S=\\\\left\\\\{${texFractionReduite(a, c - b)}\\\\right\\\\}$. `\n } else {\n correction += `The equation is of the form $\\\\dfrac{1}{x}=k$ with $k=${texFractionReduite(c - b, a)}$. So the equation admits a solution:$\\\\dfrac{1}{${texFractionReduite(c - b, a)}}=${texFractionReduite(a, c - b)}$.<br>Thus $S=\\\\left\\\\{${texFractionReduite(a, c - b)}\\\\right\\\\}$. `\n }\n }\n break\n }\n break\n\n case 'x^3=k':\n\n switch (sousChoix[i]) { // sousChoix[i] = randint(0, 5)\n case 0:\n k1 = choice([-10, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 10])\n k = k1 ** 3\n\n enonce = `Solve in $\\\\mathbb{R}$:<br>${sp(50)} $x^3=${k}$`\n correction = ''\n if (this.correctionDetaillee) {\n correction += 'For any real $k$, the equation $x^3=k$ admits as unique solution the number whose cube is equal to $k$. <br>We can note this number: $\\\\sqrt[3]{k}$. <br>'\n }\n correction += `The equation is of the form $x^3=k$ with $k=${k}$. <br>The number whose cube is $${k}$ is $${k1}$ because $${ecritureParentheseSiNegatif(k1)}^3=${k}$.<br>Thus, $S=\\\\{${k1}\\\\}$. `\n\n break\n\n case 1:\n b = randint(-10, 10, 0)\n k1 = choice([-10, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 10])\n k = k1 ** 3\n c = k + b\n enonce = `Solve in $\\\\mathbb{R}$:<br>${sp(50)} $x^3${ecritureAlgebrique(b)}=${c}$`\n correction = ''\n if (b > 0) {\n correction += `We isolate $x^3$ on the left side to obtain an equation of the type $x^3=k$.<br>$\\\\begin{aligned}x^3${ecritureAlgebrique(b)}&=${c}\\\\\\\\x^3${ecritureAlgebrique(b)}-${miseEnEvidence(b)}& =${c}-${miseEnEvidence(b)}\\\\\\\\x^3&=${c - b}\\\\end{aligned}$<br>`\n } else {\n correction += `We isolate $x^3$ on the left side to obtain an equation of the type $x^3=k$.<br>$\\\\begin{aligned}x^3${ecritureAlgebrique(b)}&=${c}\\\\\\\\x^3${ecritureAlgebrique(b)}+${miseEnEvidence(-b)}& =${c}+${miseEnEvidence(-b)}\\\\\\\\x^3&=${c - b}\\\\end{aligned}$<br>`\n }\n correction += `The equation is of the form $x^3=k$ with $k=${k}$. <br>The number whose cube is $${k}$ is $${k1}$ because $${ecritureParentheseSiNegatif(k1)}^3=${k}$.<br>Thus, $S=\\\\{${k1}\\\\}$. `\n\n break\n case 2:\n a = randint(-10, 10, [0, -1, 1])\n k1 = choice([-10, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 10])\n k = k1 ** 3\n c = k * a\n enonce = `Solve in $\\\\mathbb{R}$:<br>${sp(50)} $${a}x^3=${c}$`\n correction = ''\n\n correction += `We isolate $x^3$ on the left side to obtain an equation of the type $x^3=k$.<br>$\\\\begin{aligned}${a}x^3&=${c}\\\\\\\\x^3&=${texFractionReduite(c, a)} \\\\\\\\\\\\end{aligned}$<br>`\n\n correction += `The equation is of the form $x^3=k$ with $k=${k}$. <br>The number whose cube is $${k}$ is $${k1}$ because $${ecritureParentheseSiNegatif(k1)}^3=${k}$.<br>Thus, $S=\\\\{${k1}\\\\}$. `\n break\n case 3:\n a = randint(-10, 10, [0, -1, 1])\n b = randint(-10, 10, [0, -1, 1])\n k1 = choice([-10, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 10])\n k = k1 ** 3\n c = k * a + b\n enonce = `Solve in $\\\\mathbb{R}$:<br>${sp(50)} $${a}x^3${ecritureAlgebrique(b)}=${c}$`\n correction = ''\n\n if (b > 0) {\n correction += `We isolate $x^3$ on the left side to obtain an equation of the type $x^3=k$.<br>$\\\\begin{aligned}${a}x^3${ecritureAlgebrique(b)}&=${c}\\\\\\\\${a}x^3${ecritureAlgebrique(b)}-${miseEnEvidence(b)}& =${c}-${miseEnEvidence(b)}\\\\\\\\${a}x^3&=${texNombre(c - b, 0)}\\\\\\\\x^3&=${texFractionReduite(c - b, a)}\\\\\\\\\\\\end{aligned}$<br>`\n } else {\n correction += `We isolate $x^3$ on the left side to obtain an equation of the type $x^3=k$.<br>$\\\\begin{aligned}${a}x^3${ecritureAlgebrique(b)}&=${c}\\\\\\\\${a}x^3${ecritureAlgebrique(b)}+${miseEnEvidence(-b)}& =${c}+${miseEnEvidence(-b)}\\\\\\\\${a}x^3&=${texNombre(c - b, 0)}\\\\\\\\x^3&=${texFractionReduite(c - b, a)}\\\\\\\\\\\\end{aligned}$<br>`\n }\n\n correction += `The equation is of the form $x^3=k$ with $k=${k}$. <br>The number whose cube is $${k}$ is $${k1}$ because $${ecritureParentheseSiNegatif(k1)}^3=${k}$.<br>Thus, $S=\\\\{${k1}\\\\}$. `\n break\n }\n break\n }\n\n texte = enonce\n texteCorr = correction\n\n if (this.questionJamaisPosee(i, listeTypeDeQuestions[i], x, y, sousChoix[i])) {\n // If the question has never been asked, we create another one\n this.listeQuestions.push(texte)\n this.listeCorrections.push(texteCorr)\n i++\n }\n cpt++\n }\n listeQuestionsToContenu(this)\n }\n this.besoinFormulaireNumerique = [\n 'Choice of questions',\n 5,\n '1: x^2=k\\n2: sqrt{x}=k \\n3: 1/x=k \\n4: x^3=k \\n5: Combination'\n ]\n this.besoinFormulaire2Numerique = ['Choice of questions', 3, '1: Direct equation\\n2: Indirect equation\\n3: 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