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Binciuionutcristian
a fost răspuns

1) Rezolvati ecuatiile :
a) 108:x=18
b)72:x+2=6



2)Rezolvati in multimea numerelor naturale inecuatia:
3(x+1)<137-131

Răspuns :

Mirelarianna
108:x=18 x=108:18 x=6 72:x+2=6 72:x=6-2=4 x=74:4 x=18 3(x+1)<137-131 3(x+1)<6 x+1<6:3 x+1<2 x<2-1 x<1
Radudianaandree
a) 108:x=18
x=108:18
x=6
B) 72:x+2=6
72:x=6-2
72:x=4
x=72:4
x=18
c) 3(x+1)<137-131
3(x+1)<6
x+1<6:3
x+1<2
x=0 

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