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Care dintre ecuatile urmatoare au ca solutie un nr nat mai are decat 4 dar mai mic decat 20: 1)35·2+x=100 2)x+777-7·11=400·2-37 3)429-x=82·5 4)68·3+2·(153-71)=x+364

Răspuns :

1) x=100-35*2=100-70=30 
2)x+777-7*11=400*2-37 
=x+777-77=400*2-37
=x+700=800-37
=x+700=763
x=763-700=63
3)429-x=82*5 
=429-x=410
x=429-410=19
4)68*3+2*(153-71)=x+364
68*3+2*82=x+364
204+164=x+364
368=x+364
x=368-364=4

raspuns corect : 3)