Question

folosind reguli de calcul cu puteri calculati a) 3 la 40•3 la 18 b) 7•7 la 23:24 c) 11 la 50-11•11 la 49 d) 2 la 17•2 la 18-2 la 34•2 e) (29-2•7) la 2•15 la 20- 15 la 22 f) (12-3) la 2•(6+3) la 18-9 la 20 g) (4 la 2:2 la 2) la 2•4 la 18:256 la 5. Va rog ajutati-ma.

Answer 1

   
[tex]a) 3^{40}\times3^{18} = 3^{40+18}=3^{58} \\ b) 7\times 7^{23}:7^{24}= 7^{23+1}:7^{24}=7^{24}:7^{24}=7^{24-24}=7^0=1\\ c) 11^{50}-11\times11^{49}=11^{50}-11^{49+1}=11^{50}-11^{50} =0\\ d) 2^{17}\times2^{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 [/tex]

[tex]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) (4^2:2^2)^2 \times 4^{18}:256^5= \\ =((2^2)^2:2^2)^2 \times (2^2)^{18}:(2^8)^5= \\ =(2^{2\times2}:2^2)^2 \times 2^{2\times18}:2^{8\times 5}= \\ =(2^4:2^2)^2 \times 2^{2\times18}:2^{8\times 5}= \\ =(2^{4-2})^2 \times 2^{2\times18}:2^{8\times 5}= \\ =(2^2)^2 \times 2^{2\times18}:2^{8\times 5}= \\ =2^{2\times2} \times 2^{2\times18}:2^{8\times 5}= [/tex]

$=2^{4} \times 2^{36}:2^{40}= 2^{4+36-40}=2^{40-40}=2^0=1$