Yonelakiss99id
a fost răspuns

La un magazin s-a adus marfa. In prima zi s-a vandut 1/3 in cantitate,a doua zi s-a vandut 1/4 din rest iar in a treia zi s-a vandut 1/5 din noul rest. Stiind ca au mai ramas 480 kg de marfa ,aflati ce cantitate era la inceput.

Răspuns :

Miky93id
x-cantittatea initiala

1/3 din x= 1/3 *x= x/3-in 1 zi

x-x/3= (3x-x)/3= 2x/3-restul

1/4 din 2x/3= 1/4 * 2x/3= x/6 -in a2a zi

2x/3 -x/6= (4x-x)/6=  3x/6 -noul rest

1/5 din 3x/6= 1/5 * 3x/6= 1/5 * x/2=  x/10-in a3a zi

x/3 + x/6 + x/10 +480= x aducem la acelasi numitor
10x +5x+3x+14400=30x
18x + 14400=30x
18x-30x=-14400
-12x=-14400
12x=14400
x=1200

1/3x+ 1/4 (x-1/3x) +1/5{x-[1/3x+1/4(x-1/3x)]}+480=x

Luam fiecare paranteza si o rezolvam

1/4(x-x/3)=1/4*2x/3=2x-12=x/6 (prima paranteza)

x-[1x/3 +1/4 (x-1/3x)]=x-(1x/3+x/4-x/12)=x-[(4x+3x-x)/12]=x-6x/12=x-x/2=(2x-x)/2=x/2   (a 2-a paranteza)

Obtinem 1x/3+x/6+1/5*x/2+480=x⇒x/3+x/6+x/10+480=x⇒x/3+x/6+x/10-x=-480⇒(amplificam ca sa avem numitor comun 60)⇒(20x+10x+6x-60x)/60=-480⇒
20x+10x+6x-60x=-480*60⇒36-60x=-28800⇒-24x=-28800(se anuleaza minus cu minus)⇒24x=28800⇒x=1200 Kg



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