import{E as T,aj as N,r as $,al as Q,bc as C,aK as a,p as D,l as y}from"./index-ajJ0B2-K.js";import{R as M}from"./Relatif-YKH2MTwh.js";const L=!0,B="qcm",G=!0,W="qcmMono",I="Relative multiplications and quotients: sign with a letter",j="73187",z="4C10-6";function F(){T.call(this),this.sup=3,this.consigne="",this.correctionDetailleeDisponible=!0,this.correctionDetaillee=!1,this.spacing=2,this.nbQuestions=3,this.nbQuestionsModifiable=!0,this.nouvelleVersion=function(){this.autoCorrection=[],this.sup=parseInt(this.sup),this.listeQuestions=[],this.listeCorrections=[];let d;switch(this.sup){case 1:d=[1];break;case 2:d=[2];break;case 3:d=[1,2];break;case 4:d=[3,4];break;case 5:d=[1,2,3,4];break}const x=N(d,this.nbQuestions);for(let h=0,c,i,l,k,v,S,w=0;h<this.nbQuestions&&w<50;){this.autoCorrection[h]={};const m=20,t=new M($(-1,1,[0])*$(1,m),$(-1,1,[0])*$(1,m),$(-1,1,[0])*$(1,m),$(-1,1,[0])*$(1,m),$(-1,1,[0])*$(1,m)),R=["n","x","y","a","m"],r=R[$(0,R.length-1)],o=Q(h+1),u=$(-1,1,[0]),p=x[h]===1?$(3,5):$(4,6);let P=$(0,p-1);const e=t.relatifs.slice(0,p-1),g=[];for(let s=0;s<e.length;s++)g.push(C(e[s]));g.splice(P,0,r);let n="",b,f;switch(c=`Give the sign of $${r}$ so that ${o} is ${u===-1?"negative":"positive"}. <br>`,i=`${a("suppose that"+r+" is positive:")}`,x[h]){case 1:n+=`${g[0]}`;for(let s=1;s<p;s++)n+=`\\times ${g[s]}`;c+=` ${o} = $${n}$ <br>`,this.correctionDetaillee?(e.push(1),i+=`<br> ${t.setRegleSigneProduit(...e)}`,i+=`<br><br> So if ${a(r+" is positive","black")} $ ${n} $ is ${a(t.getSigneProduitString(...e),"black")}.`,i+=`<br><br> ${a("Now suppose that"+r+" or negative:")}`,e.push(-1),i+=`<br><br> ${t.setRegleSigneProduit(...e)}`,i+=`<br><br> So if ${a(r+" is negative","black")} $ ${n} $ is ${a(t.getSigneProduitString(...e),"black")}.`,i+=`<br><br> ${a("Conclusion :")} <br>`+a(`It is therefore necessary that $ ${r} $ be ${u===t.getSigneProduitNumber(...e)?"negative":"positive"} for ${o} to be ${u===-1?"negative":"positive"}`,"black")):(i=`$${r}$ must be ${u===t.getSigneProduitNumber(...e)?"positive":"negative"} for ${o} to be ${u===-1?"negative":"positive"}.`,S=u===t.getSigneProduitNumber(...e)?"positive":"negative");break;case 2:n+="\\dfrac {"+g[0],k=$(2,p-2);for(let s=1;s<k+1;s++)n+=`\\times ${g[s]}`;n+="}{"+g[k+1];for(let s=k+2;s<p;s++)n+=`\\times ${g[s]}`;n+="}",c+=` ${o} = $${n}$ <br>`,this.correctionDetaillee?(i+=`<br> ${t.setRegleSigneQuotient(...e)}`,i+=`<br><br> So if ${a(r+" is positive","black")} $ ${n} $ is ${a(t.getSigneProduitString(...e),"black")}.`,i+=`<br><br> ${a("Now suppose that"+r+" or negative:")}`,e.push(-1),i+=`<br> ${t.setRegleSigneQuotient(...e)}`,i+=`<br><br> So if ${a(r+" is negative","black")} $ ${n} $ is ${a(t.getSigneProduitString(...e),"black")}.`,i+=`<br><br> ${a("Conclusion :")} <br>`+a(`It is therefore necessary that $ ${r} $ be ${u===t.getSigneProduitNumber(...e)?"negative":"positive"} for ${o} to be ${u===-1?"negative":"positive"}`,"black")):i=`$${r}$ must be ${u===t.getSigneProduitNumber(...e)?"positive":"negative"} for ${o} to be ${u===-1?"negative":"positive"}.`,S=u===t.getSigneProduitNumber(...e)?"positive":"negative";break;case 3:b=$(-1,1,[0]),c=`Gives the sign of ${o} if $${r}$ is ${b===-1?"negative":"positive"}. <br>`,i="",l=$(1,3),P=$(0,p-1);for(let s=0;s<l;s++)g.splice(P,0,r);n+=`${g[0]}`;for(let s=1;s<p+l;s++)n+=`\\times ${g[s]}`;f=`${e[0]}`;for(let s=1;s<p-1;s++)f+=`\\times ${e[s]}`;c+=` ${o} = $${n}$ <br>`,this.correctionDetaillee?(l===1||l===3?(i+=`We find ${l+1} times the factor $ ${r} $.<br> But ${l+1} is even so their product will be positive.`,i+=`<br>The sign of the expression therefore has the sign of: $ ${f} $`,i+=`<br><br> ${t.setRegleSigneProduit(...e)}`,i+="<br><br>"+a(`So ${o} is ${t.getSigneProduitString(...e)} whatever the sign of $${r}$.`,"black")):(i+=`We find ${l+1} times the factor $${r}$. <br> But ${l+1} is odd so their product has the sign of $ ${r} $ or ${b===-1?"negative":"positive"}.`,b===-1?(i+=`<br>The sign of the expression therefore has the opposite sign to: $ ${f} $`,i+=`<br><br> ${t.setRegleSigneProduit(...e)}`,e.push(-1),i+="<br><br>"+a(`So ${o} is ${t.getSigneProduitString(...e)} when $${r}$ is ${b===-1?"negative":"positive"}.`,"black")):(i+=`<br>The sign of the expression therefore has the opposite sign to: $ ${f} $`,i+=`<br><br> ${t.setRegleSigneProduit(...e)}`,i+="<br><br>"+a(`So ${o} is ${t.getSigneProduitString(...e)} when $${r}$ is ${b===-1?"negative":"positive"}.`,"black"))),S=t.getSigneProduitString(...e)):(l===1||l===3?i=`${o} is ${t.getSigneProduitString(...e)} whatever the sign of $${r}$.<br>`:b===-1?(e.push(-1),i=`${o} is ${t.getSigneProduitString(...e)} if $${r}$ is negative.<br>`):i=`${o} is ${t.getSigneProduitString(...e)} if $${r}$ is positive.<br>`,S=t.getSigneProduitString(...e));break;case 4:b=$(-1,1,[0]),c=`Gives the sign of ${o} if $${r}$ is ${b===-1?"negative":"positive"}. <br>`,i="",v=$(2,7),P===0?n+=g[0]+"^{"+v+"}":n+=g[0];for(let s=1;s<p;s++)s===P?n+="\\times"+g[s]+"^{"+v+"}":n+="\\times"+g[s];f=`${e[0]}`;for(let s=1;s<p-1;s++)f+=`\\times ${e[s]}`;c+=` ${o} = $${n}$ <br>`,this.correctionDetaillee?v%2===0?(i+=`We find ${v} times the factor $ ${r} $.<br> But ${v} is even so their product will be positive.`,i+=`<br>The sign of the expression therefore has the sign of: $ ${f} $`,i+=`<br><br> ${t.setRegleSigneProduit(...e)}`,i+="<br><br>"+a(`So ${o} is ${t.getSigneProduitString(...e)} whatever the sign of $${r}$.`,"black"),S=t.getSigneProduitString(...e)):(i+=`We find ${v} times the factor $${r}$. <br> But ${v} is odd so their product has the sign of $ ${r} $ or ${b===-1?"negative":"positive"}.`,b===-1?(i+=`<br>The sign of the expression therefore has the opposite sign to: $ ${f} $`,i+=`<br><br> ${t.setRegleSigneProduit(...e)}`,e.push(-1),i+="<br><br>"+a(`So ${o} is ${t.getSigneProduitString(...e)} when $${r}$ is ${b===-1?"negative":"positive"}.`,"black")):(i+=`<br>The sign of the expression therefore has the opposite sign to: $ ${f} $`,i+=`<br><br> ${t.setRegleSigneProduit(...e)}`,i+="<br><br>"+a(`So ${o} is ${t.getSigneProduitString(...e)} when $${r}$ is ${b===-1?"negative":"positive"}.`,"black")),S=t.getSigneProduitString(...e)):(v%2===0?i=`${o} is ${t.getSigneProduitString(...e)} whatever the sign of $${r}$.<br>`:b===-1?(e.push(-1),i=`${o} is ${t.getSigneProduitString(...e)} if $${r}$ is negative.<br>`):i=`${o} is ${t.getSigneProduitString(...e)} if $${r}$ is positive.<br>`,S=t.getSigneProduitString(...e));break}this.autoCorrection[h]={enonce:c,options:{ordered:!0},propositions:[{texte:"negative",statut:S==="negative"},{texte:"zero",statut:!1},{texte:"positive",statut:S==="positive"}]},c+=D(this,h).texte,this.questionJamaisPosee(h,x[h],...e)&&(this.listeQuestions.push(c),this.listeCorrections.push(i),h++),w++}y(this)},this.besoinFormulaireNumerique=["Difficulty level",5,`1: Multiplications
2: Quotients 
3: Multiplications and quotients 
4: Multiplications with several times the letter (including powers) 
5: Mixture`]}export{G as amcReady,W as amcType,F as default,L as interactifReady,B as interactifType,z as ref,I as titre,j as uuid};
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