👤
a fost răspuns

efectuati:(am sa pun ghilimele in locul puterilor deoarece nu-mi merge tastatura bine)a)3''40:[3''20x3''18+(3''5x3''14)''2+(4''20:4''19-1")''38];b)8''33:[4''32x2''34+(2''5x2''20)''5:(16x2''23)+(75:75-2''0)x32x4];c)[(34)2+26:22-20]:[(32)4-42-30];

Răspuns :

Renatemambouko
a)3^40:[3^20x3^18+(3^5x3^14)^2+(4^'20:4^19-1)^38]=
=3^40:[3^38+(3^19)^2+(2^40:2^38-1)^38]=
=3^40:[3^38+3^38+(2^2-1)^38]=
=3^40:[3^38(3^0+3^0+3^0]=
=3^40:3^38*3=
=3^40:3^39=
=3

b)8^33:[4^32x2^34+(2^5x2^20)5:(16x2^23)+(75:75-2^0)x32x4]
=2^99:[2^64x2^34+(2^25)5:(2^4x2^23)+(1-1)x32x4]=
=2^99:[2^98+2^125:(2^27)+0]=
=2^99:(2^98+2^98)=
=2^99:2^98(2^0+2^0)=
=2^99:2^98*2=
=2^99:2^99=1





























Alte intrebari