import{E as T,aj as C,r as b,D as m,W as y,w as s,x as $,a as g,m as i,s as o,g as v,l as E}from"./index-XCg2QAX4.js";const D="Framing with square roots",W=!0,A="mathLive",L="28/09/2022",N="ed2ee",R="2N12-3";function j(){T.call(this),this.nbQuestions=2,this.nbCols=1,this.nbColsCorr=1,this.sup=1,this.nouvelleVersion=function(){this.listeQuestions=[],this.listeCorrections=[],this.autoCorrection=[],this.sup=parseInt(this.sup);let d;this.sup===1?d=["Frame1"]:this.sup===2?d=["Frame2"]:this.sup===3?d=["Frame3"]:this.sup===4&&(d=["Frame1","Frame2","Frame3"]);const x=C(d,this.nbQuestions);for(let a=0,t,r,e,q,h,F,p,w,n,l,f,u,M,c,k=0;a<this.nbQuestions&&k<50;){switch(x[a]){case"Frame1":t=b(3,143,[4,9,16,25,36,49,64,81,100,121]),F=Math.floor(Math.sqrt(t)),q=`Enclose $\\sqrt{${t}}$ by two consecutive integers.<br>`,this.interactif&&(q+=g(this,2*a,"largeur10 inline")+`$< \\sqrt{${t}} <$`+g(this,2*a+1,"largeur10 inline nospacebefore")),h=`As $${t}$ is not the square of an integer, we frame $${t}$ by two squares of integers: <br>$${Math.floor(Math.sqrt(t))**2} < ${t} < ${(Math.floor(Math.sqrt(t))+1)**2}$, i.e. $${Math.floor(Math.sqrt(t))}^2 < ${t} < ${Math.floor(Math.sqrt(t))+1}^2 $.<br><br>Taking the square root of each of these numbers, we obtain: <br>$\\sqrt{${Math.floor(Math.sqrt(t))}^2} < \\sqrt{${t}} < \\sqrt{${Math.floor(Math.sqrt(t))+1}^2} $(we do not change the direction of the inequalities by taking the square roots. This accepted result will be demonstrated in the chapter on variations). <br><br>Finally, we obtain the framing of $\\sqrt{${t}}$ by two consecutive integers: $${Math.floor(Math.sqrt(t))}< \\sqrt{${t}} < ${Math.floor(Math.sqrt(t))+1}$. `,o(this,2*a,F),o(this,2*a+1,F+1);break;case"Frame2":t=b(3,143,[4,9,16,25,36,49,64,81,100,121]),r=b(-9,9,0),e=b(-9,9,[0,1]),p=r+e*Math.floor(Math.sqrt(t)),w=r+e*Math.floor(Math.sqrt(t)+1),q=`Using a bracket of $\\sqrt{${t}}$ by two consecutive integers, give a bracket of $${r}${$(e)}\\sqrt{${t}}$ as precise as possible.<br>`,this.interactif&&(q+=g(this,2*a,"largeur10 inline")+`$< ${r}${$(e)}\\sqrt{${t}} <$`+g(this,2*a+1,"largeur10 inline nospacebefore")),h=`As $${t}$ is not the square of an integer, we frame $${t}$ by two squares of integers: <br>$${Math.floor(Math.sqrt(t))**2} < ${t} < ${(Math.floor(Math.sqrt(t))+1)**2}$, i.e. $${Math.floor(Math.sqrt(t))}^2 < ${t} < ${Math.floor(Math.sqrt(t))+1}^2 $.<br><br>Taking the square root of each of these numbers, we obtain: <br>$\\sqrt{${Math.floor(Math.sqrt(t))}^2} < \\sqrt{${t}} < \\sqrt{${Math.floor(Math.sqrt(t))+1}^2} $(we do not change the direction of the inequalities by taking the square roots. This accepted result will be demonstrated in the chapter on variations). <br><br>Finally, we obtain the framing of $\\sqrt{${t}}$ by two consecutive integers: $${Math.floor(Math.sqrt(t))}< \\sqrt{${t}} < ${Math.floor(Math.sqrt(t))+1}$. <br><br>Starting from this framework, we successively obtain:<br>`,e>0?(h+=`$\\begin{aligned}${Math.floor(Math.sqrt(t))} &< \\sqrt{${t}} < ${Math.floor(Math.sqrt(t))+1}\\\\${i(e)}\\times ${Math.floor(Math.sqrt(t))}&< ${i(e)}\\times \\sqrt{${t}} < ${i(e)}\\times ${Math.floor(Math.sqrt(t))+1}{\\text{ ( We multiply by a strictly positive number)}}\\\\${e*Math.floor(Math.sqrt(t))}&< ${e}\\sqrt{${t}} < ${e*(Math.floor(Math.sqrt(t))+1)}\\\\${i(r)}${v(e*Math.floor(Math.sqrt(t)))}&< ${i(r)}${$(e)}\\sqrt{${t}} < ${i(r)}${v(e*(Math.floor(Math.sqrt(t))+1))}\\\\${r+e*Math.floor(Math.sqrt(t))}&< ${r}${$(e)}\\sqrt {${t}} < ${r+e*(Math.floor(Math.sqrt(t))+1)}\\end{aligned}$<br>The requested framing is therefore: $ ${r+e*Math.floor(Math.sqrt(t))}< ${r}${$(e)}\\sqrt{${t}} < ${r+e*(Math.floor(Math.sqrt(t))+1)}$. `,o(this,2*a,p),o(this,2*a+1,w)):(h+=`$\\begin{aligned}${Math.floor(Math.sqrt(t))} &< \\sqrt{${t}} < ${Math.floor(Math.sqrt(t))+1}\\\\${i(e)}\\times ${Math.floor(Math.sqrt(t))}&> ${i(e)}\\times \\sqrt{${t}} > ${i(e)}\\times ${Math.floor(Math.sqrt(t))+1}{\\text{ ( We multiply by a strictly negative number)}}\\\\${e*Math.floor(Math.sqrt(t))}&> ${e}\\sqrt{${t}} > ${e*(Math.floor(Math.sqrt(t))+1)}\\\\${i(r)}${v(e*Math.floor(Math.sqrt(t)))}&> ${i(r)}${$(e)}\\sqrt{${t}} > ${i(r)}${v(e*(Math.floor(Math.sqrt(t))+1))}\\\\${r+e*Math.floor(Math.sqrt(t))}&> ${r}${$(e)}\\sqrt {${t}} > ${r+e*(Math.floor(Math.sqrt(t))+1)}\\end{aligned}$<br>The requested framing is therefore: $ ${r+e*(Math.floor(Math.sqrt(t))+1)}< ${r}${$(e)}\\sqrt{${t}} < ${r+e*Math.floor(Math.sqrt(t))}$. `,o(this,2*a,w),o(this,2*a+1,p));break;case"Frame3":t=b(3,143,[4,9,16,25,36,49,64,81,100,121]),r=b(-9,9,0),e=b(-9,9,[0,1]),n=new m(y(Math.sqrt(t)-.05,1)),u=new m(n).mul(e),f=new m(y(u)).add(r),l=new m(y(Math.sqrt(t)+.05,1)),c=new m(l).mul(e),M=new m(y(c)).add(r),p=t,w=t,q=`Using the framework $${s(n,1)}<\\sqrt{${t}}<${s(l,1)}$, give a framework of $${r}${$(e)}\\sqrt{${t}}$ as precise as possible.<br>`,this.interactif&&(q+=g(this,2*a,"largeur10 inline")+`$< ${r}${$(e)}\\sqrt{${t}} <$`+g(this,2*a+1,"largeur10 inline nospacebefore")),h=`From the framework $${s(n,1)}<\\sqrt{${t}}<${s(l,1)}$, we successively obtain:<br>`,e>0?(h+=`$\\begin{aligned}${s(n,1)} &< \\sqrt{${t}} < ${s(l,1)}\\\\${i(e)}\\times ${s(n,1)}&< ${i(e)}\\times \\sqrt{${t}} < ${i(e)}\\times ${s(l,1)}{\\text{ ( We multiply by a strictly positive number)}}\\\\${s(u,1)}&< ${e}\\sqrt{${t}} <${s(c,1)}\\\\${i(r)}+${s(u,1)}&< ${i(r)}${$(e)}\\sqrt{${t}} < ${i(r)}+${s(c,1)}\\\\${s(f,1)}&< ${r}${$(e)}\\sqrt{${t}} < ${s(M,1)}\\end{aligned}$<br>The requested framing is therefore: $ ${s(f,1)}< ${r}${$(e)}\\sqrt{${t}} < ${s(M,1)}$. `,o(this,2*a,f),o(this,2*a+1,M)):(h+=`$\\begin{aligned}${s(n,1)} &< \\sqrt{${t}} < ${s(l,1)}\\\\${i(e)}\\times ${s(n,1)}&> ${i(e)}\\times \\sqrt{${t}} > ${i(e)}\\times ${s(l,1)}{\\text{ ( We multiply by a strictly negative number)}}\\\\${s(u,1)}&> ${e}\\sqrt{${t}} >${s(c,1)}\\\\${i(r)}${s(u,1)}&> ${i(r)}${$(e)}\\sqrt{${t}} > ${i(r)}${s(c,1)}\\\\${s(f,1)}&> ${r}${$(e)}\\sqrt {${t}} > ${s(M,1)}\\end{aligned}$<br>The requested framing is therefore: $ ${s(M,1)}< ${r}${$(e)}\\sqrt{${t}} < ${s(f,1)}$. `,o(this,2*a,M),o(this,2*a+1,f));break}this.questionJamaisPosee(a,t,r,e)&&(this.listeQuestions.push(q),this.listeCorrections.push(h),a++),k++}E(this)},this.besoinFormulaireNumerique=["Choice of questions",4,`1: Frame sqrt(a)
2: Frame a+b*sqrt(c) with integers
3: Frame a+b*sqrt(c) with decimals
4: Mix`]}export{L as dateDePublication,j as default,W as interactifReady,A as interactifType,R as ref,D as titre,N as uuid};
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