Question

11 Folosind reguli de calcul cu puteri , calculaţi : b) 7•7^23÷7^24
c) 11^50-11•11^49
d) 2^17•2^18•2^34•2
e) (29-2•7)^2•15^20-15^22
f) (12-3)^2•(6+3)^18-9^20
g) (4^2÷3^2)•4^18÷256^5


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Va rog.Urgentt.Dau coroana
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Answer 1

Explicație pas cu pas:

b)

$7 \times {7}^{23} \div {7}^{24} = {7}^{1 + 23} \div {7}^{24} = {7}^{24} \div {7}^{24} = 1$

c)

${11}^{50} - 11 \times {11}^{49} = {11}^{50} - {11}^{1 + 49} = {11}^{50} - {11}^{50} = 0$

d)

${2}^{17} \times {2}^{18} - {2}^{34} \times 2 = {2}^{17 + 18} - {2}^{34 + 1} = {2}^{35} - {2}^{35} = 0$

e)

${(29 - 2 \times 7)}^{2} \times {15}^{20} - {15}^{22} =$

$= {(29 - 14)}^{2} \times {15}^{20} - {15}^{22} =$

$= {15}^{2} \times {15}^{20} - {15}^{22} = {15}^{2 + 20} - {15}^{22}$

$= {15}^{22} - {15}^{22} = 0$

f)

${(12 - 3)}^{2} \times {(6 + 3)}^{18} - {9}^{20}=$

$= {9}^{2} \times {9}^{18} - {9}^{20} = {9}^{2 + 18} - {9}^{20}$

$= {9}^{20} - {9}^{20} = 0$

g)

$(4^{2}:3^{2})*4^{18}:256^{5} = (4^{2+18}:3^{2}):(4^{4})^{5} = (4^{20}:3^{2}):(4)^{20} = 4^{20-20}:3^{2} = 4^{0}:3^{2} = 1:3^{2} = 3^{-2} = \frac{1}{9}$