import{E as d,h as q,aj as p,r as o,j as t,aq as b,ay as a,l as y}from"./index-XCg2QAX4.js";const F="Test if a number is the solution to an equation",W="a1c9a",Q="4L14-0";function k(){d.call(this),this.titre=F,this.consigne="",this.nbCols=1,this.nbColsCorr=1,this.sup=1,this.exo==="4L14-1"?this.nbQuestions=4:this.exo==="4L14-2"?this.nbQuestions=3:this.nbQuestions=9,this.nouvelleVersion=function(){this.listeQuestions=[],this.listeCorrections=[],this.sup=Number(this.sup);let f;this.exo==="4L14-1"?f=[q([1,2]),3,q([4,5]),8]:this.exo==="4L14-2"?f=[9,6,7]:f=[1,2,3,4,5,8,6,7,9];const c=p(f,this.nbQuestions);this.consigne="Justify whether the proposed numbers are solutions of the given equation or not.";for(let m=0,u,i,l=0;m<this.nbQuestions&&l<50;){let $,e,x,h,s,r,n;switch(c[m]){case 1:this.sup===1?($=o(1,6),e=o(1,6,[$]),r=$+e,s=o(2,10,[r])):($=o(-6,6,[0]),e=o(-6,6,[$,0]),r=$+e,s=o(-10,10,[0,r])),u=`$3x-${t($)}=2x+${t(e)}~$ for $~x=${s}~$ then for $~x=${r}$`,i=`For $x=${s}$: <br>`,i+=`$3x-${t($)}=3\\times ${t(s)}-${t($)}=${3*s-$}$ <br> $2x+${t(e)}=2\\times ${t(s)}+${t(e)}=${2*s+e}$<br>`,i+=`$${3*s-$}
=${2*s+e}$ so the equality is not true.<br>`,i+=`${b(`$x=${s}$ is therefore not a solution to the equation $3x-${t($)}=2x+${t(e)}~$`)}<br><br>`,i+=`For $x=${t(r)}$: <br>`,i+=`$3x-${t($)}=3\\times ${t(r)}-${t($)}=${3*r-$}$ <br> $2x+${t(e)}=2\\times ${t(r)}+${t(e)}=${2*r+e}$<br>`,i+="We find the same result for the left member and for the right member so the equality is true.<br>",i+=`${b(`$x=${r}$ is therefore solution of the equation $3x-${t($)}=2x+${t(e)}~$`)}`;break;case 2:this.sup===1?($=o(1,9),e=o(0,4)*2+$%2,s=parseInt(a(($+e)/2)),r=o(1,9,s)):($=o(-9,9,[0]),e=o(-4,4,[$,0])*2+$%2,s=parseInt(a(($+e)/2)),r=o(-9,9,[0,s])),u=`$3x+${t($)}=5x-${t(e)}~$ for $~x=${s}~$ then for $~x=${r}$`,i=`For $x=${s}$: <br>`,i+=`$3x+${t($)}=3\\times ${t(s)}+${t($)}=${3*s+$}$ <br> $5x-${t(e)}=5\\times ${t(s)}-${t(e)}=${5*s-e}$<br>`,i+="We find the same result for the left member and for the right member so the equality is true.<br>",i+=`${b(`$x=${s}$ is therefore solution of the equation $3x+${t($)}=5x-${t(e)}~$`)}<br><br>`,i+=`For $x=${r}$: <br>`,i+=`$3x+${t($)}=3\\times ${t(r)}+${t($)}=${3*r+$}$ <br> $5x-${t(e)}=5\\times ${t(r)}-${t(e)}=${5*r-e}$<br>`,i+=`$${3*r+$}
=${5*r-e}$ so the equality is not true.<br>`,i+=`${b(`$x=${r}$ is therefore not a solution to the equation $3x+${t($)}=5x-${t(e)}~$`)}`;break;case 3:this.sup===1?($=o(1,3),e=o(1,3),r=parseInt(a((10*$+4*e)/2)),s=o(1,9,r)):($=o(-3,3,[0]),e=o(-3,3,[0]),r=parseInt(a((10*$+4*e)/2)),s=o(-9,9,[0,r])),u=`$10(x-${t($)})=4(2x+${t(e)})~$ for $~x=${s}~$ then for $~x=${r}$`,i=`For $x=${s}$: <br>`,i+=`$10(x-${t($)})=10\\times (${t(s)}-${t($)})=10\\times ${s-$}=${10*(s-$)}$ <br> $4(2x+${t(e)})=4\\times (2\\times ${t(s)}+${t(e)})=4 \\times ${2*s+e}=${4*(2*s+e)}$<br>`,i+=`$${10*(s-$)}
=${4*(2*s+e)}$ so the equality is not true.<br>`,i+=`${b(`$x=${s}$ is therefore not a solution to the equation $10(x-${t($)})=4(2x+${t(e)})~$`)}<br><br>`,i+=`For $x=${r}$: <br>`,i+=`$10(x-${t($)})=10\\times (${t(r)}-${t($)})=10\\times ${r-$}=${10*(r-$)}$ <br> $4(2x+${t(e)})=4\\times (2\\times ${t(r)}+${t(e)})=4 \\times ${2*r+e}=${4*(2*r+e)}$<br>`,i+="We find the same result for the left member and for the right member so the equality is true.<br>",i+=`${b(`$x=${r}$ is therefore solution of the equation $10(x-${t($)})=4(2x+${t(e)})~$`)}`;break;case 4:this.sup===1?($=o(2,9),e=o(2,9),x=o(1,3),s=e+x,r=o(2,10,s)):($=o(2,9),e=o(2,9)*o(-1,1,0),x=o(1,3)*o(-1,1,0),s=e+x,r=o(2,10,s)*o(-1,1,0)),u=`$${t($)}x+${t(e)}=${$+1}x-${t(x)}~$ for $~x=${s}~$ then for $~x=${r}$`,i=`For $x=${s}$: <br>`,i+=`$${$}x+${t(e)}=${t($)}\\times ${t(s)}+${t(e)}=${$*s+e}$ <br> $${$+1}x-${t(x)}=${$+1}\\times ${t(s)}-${t(x)}=${($+1)*s-x}$<br>`,i+="We find the same result for the left member and for the right member so the equality is true.<br>",i+=`${b(`$x=${s}$ is therefore solution of the equation $${t($)}x+${t(e)}=${$+1}x-${t(x)}~$`)}<br><br>`,i+=`For $x=${r}$: <br>`,i+=`$${$}x+${t(e)}=${t($)}\\times ${t(r)}+${t(e)}=${$*r+e}$ <br> $${$+1}x-${t(x)}=${$+1}\\times ${t(r)}-${t(x)}=${($+1)*r-x}$<br>`,i+=`$${$*r+e}
=${($+1)*r-x}$ so the equality is not true.<br>`,i+=`${b(`$x=${r}$ is therefore not a solution to the equation $${t($)}x+${t(e)}=${$+1}x-${t(x)}~$`)}<br><br>`;break;case 5:this.sup===1?(s=o(1,9),e=o(1,9),$=e+4*s,r=o(1,11,s)):(s=o(-9,9),e=o(-9,9,0),$=e+4*s,r=o(1,11,s)),u=`$${$}-2x=${e}+2x~$ for $~x=${s}~$ then for $~x=${r}$`,i=`For $x=${s}$: <br>`,i+=`$${$}-2x=${$}-2\\times ${t(s)}=${$-2*s}$ <br> $${e}+2x=${e}+2\\times ${t(s)}=${e+2*s}$<br>`,i+="We find the same result for the left member and for the right member so the equality is true.<br>",i+=`${b(`$x=${s}$ is therefore solution of the equation $${$}-2x=${e}+2x~$`)}<br><br>`,i+=`For $x=${r}$: <br>`,i+=`$${$}-2x=${$}-2\\times ${t(r)}=${$-2*r}$ <br> $${e}+2x=${e}+2\\times ${t(r)}=${e+2*r}$<br>`,i+=`$${$-2*r}
=${e+2*r}$ so the equality is not true.<br>`,i+=`${b(`$x=${r}$ is therefore not a solution to the equation $${$}-2x=${e}+2x~$`)}<br><br>`;break;case 6:this.sup===1?(e=o(2,9),$=o(2,9,[e]),n=e,s=$,r=o(1,9,[s,n])):($=o(-9,9,[0,1]),e=o(-9,9,[0,$]),s=$,n=e,r=o(-9,9,[s,n])),u=`$${$}x-${t($*e)}=x^2-${t(e)}x~$ for $~x=${s}~$ , for $~x=${r}~$ then for $~x=${n}$`,i=`For $x=${s}$: <br>`,i+=`$${$}x-${t($*e)}=${$}\\times ${t(s)}-${t($*e)}=${$*s-$*e}$ <br> $x^2-${t(e)}\\times x=${t(s)}^2-${t(e)}\\times ${t(s)}=${s*s}-${t(e*s)}=${s*s-e*s}$<br>`,i+="We find the same result for the left member and for the right member so the equality is true.<br>",i+=`${b(`$x=${s}$ is therefore solution of the equation $${$}x-${t($*e)}=x^2-${t(e)}x~$`)}<br><br>`,i+=`For $x=${r}$: <br>`,i+=`$${$}x-${t($*e)}=${$}\\times ${t(r)}-${t($*e)}=${$*r-$*e}$ <br> $x^2-${e}\\times x=${t(r)}^2-${t(e)}\\times ${t(r)}=${r*r}-${t(e*r)}=${r*r-e*r}$<br>`,i+=`$${$*r-$*e}
=${r*r-e*r}$ so the equality is not true.<br>`,i+=`${b(`$x=${r}$ is therefore not a solution to the equation $${$}x-${t($*e)}=x^2-${t(e)}x~$`)}<br><br>`,i+=`For $x=${n}$: <br>`,i+=`$${$}x-${t($*e)}=${$}\\times ${t(n)}-${t($*e)}=${$*n-$*e}$ <br> $x^2-${e}\\times x=${t(n)}^2-${t(e)}\\times ${t(n)}=${n*n}-${t(e*n)}=${n*n-e*n}$<br>`,i+="We find the same result for the left member and for the right member so the equality is true.<br>",i+=`${b(`$x=${n}$ is therefore solution of the equation $${$}x-${t($*e)}=x^2-${t(e)}x~$`)}`;break;case 7:this.sup===1?(x=o(2,5),$=o(2,5),r=o(2,6),n=o(2,6,r),s=o(1,7,[r,n]),e=$*r,h=x*n):(x=o(2,5)*o(-1,1,0),$=o(2,5)*o(-1,1,0),r=o(1,6)*o(-1,1,0),n=o(1,6,r)*o(-1,1,0),s=o(1,7,[r,n])*o(-1,1,0),e=$*r,h=x*n),u=`$${$*h}x-${t(e*h)}=${$*x}x^2-${t(e*x)}x~$ for $~x=${s}~$, for $~x=${r}~$ then for $~x=${n}$`,i=`For $x=${s}$: <br>`,i+=`$${$*h}x-${t(e*h)}=${$*h}\\times ${t(s)}-${t(e*h)}=${$*h*s-h*e}$ <br> $${$*x}x^2-${t(e*x)}x=${$*x}\\times ${t(s)}^2-${t(e*x)}\\times ${t(s)}=${$*x*s*s}-${t(e*x*s)}=${$*x*s*s-e*x*s}$<br>`,i+=`$${$*h*s-h*e}
=${$*x*s*s-e*x*s}$ so the equality is not true.<br>`,i+=`${b(`$x=${s}$ is therefore not a solution to the equation $${$*h}x-${t(e*h)}=${$*x}x^2-${t(e*x)}x~$`)}<br><br>`,i+=`For $x=${r}$: <br>`,i+=`$${$*h}x-${t(e*h)}=${$*h}\\times ${t(r)}-${t(e*h)}=${$*h*r-h*e}$ <br> $${$*x}x^2-${t(e*x)}x=${$*x}\\times ${t(r)}^2-${t(e*x)}\\times ${t(r)}=${$*x*r*r}-${t(e*x*r)}=${$*x*r*r-e*x*r}$<br>`,i+="We find the same result for the left member and for the right member so the equality is true.<br>",i+=`${b(`$x=${r}$ is therefore solution of the equation $${$*h}x-${t(e*h)}=${$*x}x^2-${t(e*x)}x~$`)}<br><br>`,i+=`For $x=${n}$: <br>`,i+=`$${$*h}x-${t(e*h)}=${$*h}\\times ${t(n)}-${t(e*h)}=${$*h*n-h*e}$ <br> $${$*x}x^2-${t(e*x)}x=${$*x}\\times ${t(n)}^2-${t(e*x)}\\times ${t(n)}=${$*x*n*n}-${t(e*x*n)}=${$*x*n*n-e*x*n}$<br>`,i+="We find the same result for the left member and for the right member so the equality is true.<br>",i+=`${b(`$x=${n}$ is therefore solution of the equation $${$*h}x-${t(e*h)}=${$*x}x^2-${t(e*x)}x~$`)}`;break;case 8:this.sup===1?($=o(1,3),e=o(1,3),r=parseInt(a((4*$+4*e)/4)),s=o(9,r)):($=o(-3,3,[0]),e=o(-3,3,[0]),r=parseInt(a((4*$+4*e)/4)),s=o(-9,9,[0,r])),u=`$12x-${t(4*$)}=4(2x+${t(e)})~$ for $~x=${s}~$ then for $~x=${r}$`,i=`For $x=${s}$: <br>`,i+=`$12x-${t(4*$)}=12\\times ${t(s)}-${t(4*$)}=${12*s-4*$}$ <br> $4(2x+${t(e)})=4\\times (2\\times ${t(s)}+${t(e)})=4\\times ${2*s+e}=${4*(2*s+e)}$<br>`,i+=`$${12*s-4*$}
=${4*(2*s+e)}$ so the equality is not true.<br>`,i+=`${b(`$x=${s}$ is therefore not a solution to the equation $12x-${t(4*$)}=4(2x+${t(e)})~$`)}<br><br>`,i+=`For $x=${r}$: <br>`,i+=`$12x-${t(4*$)}=12\\times ${t(r)}-${t(4*$)}=${12*r-4*$}$ <br> $4(2x+${t(e)})=4\\times (2\\times ${t(r)}+${t(e)})=4\\times ${2*r+e}=${4*(2*r+e)}$<br>`,i+="We find the same result for the left member and for the right member so the equality is true.<br>",i+=`${b(`$x=${r}$ is therefore solution of the equation $12x-${t(4*$)}=4(2x+${t(e)})~$`)}<br><br>`;break;case 9:if(this.sup===1)e=o(2,9),$=o(2,9),n=e,s=$,r=o(1,9,[s,n]);else do $=o(-9,9,[0,1]),e=o(-9,9,[0,$]),s=$,n=e,r=o(-9,9,[s,n]);while($+e===0||$+e===1);u=`$x^2-${t(e+$)}x+${t($*e)}=0~$ for $~x=${s}~$ , for $~x=${r}~$ then for $~x=${n}$`,i=`For $x=${s}$: <br>`,i+=`$x^2-${t(e+$)}\\times x+${t($*e)}=${t(s)}^2-${t($+e)}\\times ${t(s)}+${t($*e)}=${s*s}-${t(($+e)*s)}+${t($*e)}=${s*s-($+e)*s+$*e}$<br>`,i+="We find $0$ for the left side so the equality is true.<br>",i+=`${b(`$x=${s}$ is therefore solution of the equation $x^2-${t(e+$)}x-${t($*e)}=0~$`)}<br><br>`,i+=`For $x=${r}$: <br>`,i+=`$x^2-${t(e+$)}\\times x+${t($*e)}=${t(r)}^2-${t($+e)}\\times ${t(r)}+${t($*e)}=${r*r}-${t(($+e)*r)}+${t($*e)}=${r*r-($+e)*r+$*e}$<br>`,i+=`$${r*r-($+e)*r+$*e}
=0$ so the equality is not true.<br>`,i+=`${b(`$x=${r}$ is therefore not a solution to the equation $x^2-${t(e+$)}x-${t($*e)}=0~$`)}<br><br>`,i+=`For $x=${n}$: <br>`,i+=`$x^2-${t(e+$)}\\times x+${t($*e)}=${t(n)}^2-${t($+e)}\\times ${t(n)}+${t($*e)}=${n*n}-${t(($+e)*n)}+${t($*e)}=${n*n-($+e)*n+$*e}$<br>`,i+="We find $0$ for the left side so the equality is true.<br>",i+=`${b(`$x=${n}$ is therefore solution of the equation $x^2-${t(e+$)}x-${t($*e)}=0~$`)}`;break}this.listeQuestions.indexOf(u)===-1&&(this.listeQuestions.push(u),this.listeCorrections.push(i),m++),l++}y(this)},this.besoinFormulaireNumerique=["Difficulty level",2,`1: Natural integers
2: Relative integers`]}export{k as default,Q as ref,F as titre,W as uuid};
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