Cum se rezolva operatia [2 la puterea a 7 totul la a 5: la 2 la puterea a 14 ori (2 la puterea a 6 plus 2 la puterea a 6 ) minus 4la a 6 : la 2 la a 10 ] :(2la puterea 28 minus 2 la puterea a 2) plus 9 va rog ajutatima rapid !

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Răspunsuri

2014-10-26T14:58:27+02:00
[(2^7)^5:2^14·(2^6+2^6)-4^6:2^10]:(2^28-2²)+9=
[2^35:2^14·2^6(1+1)-(2²)^6:2^10]:(2^28-2²)+9=
[2^(35-14)·2^6·2-2^12:2^10]:(2^28-2²)+9=
[2^21·2^(6+1)-2^(12-10)]:(2^28-2²)+9=
[2^(21+7)-2²]:(2^28-2²)+9=
[2^28-2²]:(2^28-2²)+9=1+9=10
2 3 2