12. Calculați, folosind proprietatea a" :a" = am-n:
a) 815 : 812
b) 1331 :1330 +13
c) 729 : 726 – 72
d) 717:713–73 +13°
e) (11+4)22 :1520 – 15°
A (42:14)2.918:276
Repede va rogg!!

Răspuns :

Explicație pas cu pas:

a)

[tex] {8}^{15}:{8}^{12}={8}^{15 - 12}={8}^{3}= ({2}^{3})^{3} = {2}^{3 \cdot3} = \boxed{{2}^{9}}[/tex]

b)

[tex] {13}^{31}:{13}^{30} + 13 = {13}^{31 - 30} + 13 = {13}^{1} + 13 = \boxed{26}[/tex]

c)

[tex] {7}^{29}: {7}^{26}-{7}^{2}={7}^{29 - 26} - {7}^{2} = {7}^{3}-{7}^{2} = \\\\{7}^{2} \cdot( {7}^{1} - 1)={7}^{2}\cdot 6 = 49 \cdot6 = \boxed{294}[/tex]

d)

[tex] {7}^{17}: {7}^{13}-{7}^{3} + {13}^{0} ={7}^{17 - 13}-{7}^{3} + {13}^{0}= {7}^{4}-{7}^{3} + 1= \\\\{7}^{3} \cdot( {7}^{1} -1) + 1={7}^{3}\cdot 6 + 6 = 343 \cdot6 + 1 = \boxed{2059}[/tex]

e)

[tex] (11+4)^{22}:15^{20}-{15}^{0}= 15^{22}:15^{20}-1= \\\\ 15^{22-20}-1=15^{2}-1 = 225-1 = \boxed{224} [/tex]

f)

[tex](42:14)^{2}\cdot 9^{18}:27^{6} = 3^{2}\cdot ({3}^{2})^{18}: (3^{3})^{6} = \\\\3^{2}\cdot 3^{2 \cdot18}: 3^{3 \cdot6} =3^{2}\cdot 3^{36}: 3^{18} =3^{2+36}: 3^{18} =\\\\3^{38}: 3^{18} = 3^{38-18}=\boxed{ 3^{20}}[/tex]