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Istrateadrian56
a fost răspuns

A) |-1 3supra4| × |-2|-|-1 1 supra 6 |=
B) |2supra3 -1supra4|÷|-5supra6|+(-2)×|-1supra2| la puterea a 3-a (doar ultimul modul este la puterea a 3-a ). Cu rezolvare la ambele va rogg !

Răspuns :

1intreg1/6=(1·6+1)/6=7/6
I-13/4I=13/4
I-2I=2
I-7/6I=7/6
I-13/4I·I-2I-I-7/6I=(13/4)·2-7/6=26/4-7/6 aducem la acelasi numitor=
(26·3)/12-(7·2)/12=78/12-14/12=64/12 simplif.cu 4=16/3=5,(3)

B) nu exista modul la putere
I2/3-1/4I=I8/12-3/12I=I5/12I=5/12
I-5/6I=5/6
I(-1/2)³I=I-1/8I=1/8
5/12:5/6+(-2)·1/8=5/12·6/5-2/8=simplif.=1/2-1/4=(2-1)/4=1/4=0,25


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