W załącznikach 1 i 2 chodzi o rozwiązanie zadań z potęgami. W podpunkcie a) podkreślone cyfry to JED.... Question from @jedencztery

November 2018 1 40 Report

W załącznikach 1 i 2 chodzi o rozwiązanie zadań z potęgami. W podpunkcie a) podkreślone cyfry to JEDNA potęga.

Roma

Zad. 1

$(36 \cdot 18^n -8 \cdot 2^{n-4} \cdot 9^n - 3^{n+1} \cdot 6^{n+1}) :18^{n-1} = \\\\ =(2 \cdot 18 \cdot 18^n -2^3 \cdot 2^{n-4} \cdot 9^n - (3 \cdot 6)^{n+1}) : \frac{18^n}{18^1} = \\\\ =(2 \cdot 18^{n+1} -2^{3+n-4} \cdot 9^n - 18^{n+1}) \cdot \frac{18^1}{18^n} = \\\\ =(18^{n+1} -2^{n-1} \cdot 9^n)\cdot \frac{18}{18^n} =(18^{n+1}-\frac{2^n}{2} \cdot 9^n)\cdot \frac{18}{18^n} =$

$=(18 \cdot 18^n-\frac{(2\cdot 9)^n}{2})\cdot \frac{18}{18^n}=(\frac{36 \cdot 18^n}{2}-\frac{18^n}{2})\cdot \frac{18}{18^n} = \\\\ =\frac{36 \cdot 18^n-18^n}{2}\cdot \frac{18}{18^n} =\frac{35 \cdot 18^n}{2}\cdot \frac{18}{18^n} =35 \cdot 9 =315$

$\frac{2^{19} \cdot 27^3 +15 \cdot 4^9 \cdot 9^4}{6^9 \cdot 2^{10} +12^{10}} = \frac{2^{19} \cdot (3^3)^3 +15 \cdot (2^2)^9 \cdot (3^2)^4}{(2 \cdot3)^9 \cdot 2^{10} +(2^2 \cdot 3)^{10}} =\frac{2^{19} \cdot 3^9 +5 \cdot 3 \cdot 2^{18} \cdot 3^8}{2^9 \cdot3^9 \cdot 2^{10} +2^{20} \cdot 3^{10}} = \\\\ =\frac{2^{19} \cdot 3^9 +5 \cdot 2^{18} \cdot 3^9}{2^{19} \cdot3^9+2^{20} \cdot 3^{10}} =\frac{2^{18} \cdot 3^9 \cdot (2 +5)}{2^{18} \cdot3^9 (2+2^2 \cdot 3)} = \frac{7}{14} = \frac{1}{2}$

$(10^{12} + 5^{11} \cdot 2^9 - 5^{13} \cdot 2^8) : (4 \cdot 5^5 \cdot 10^6) = \\\\ = ((2 \cdot5)^{12}+ 5^{11} \cdot 2^9 - 5^{13} \cdot 2^8) : (2^2 \cdot 5^5 \cdot (2 \cdot 5)^6) = \\\\ =(2^{12} \cdot 5^{12}+ 5^{11} \cdot 2^9 - 5^{13} \cdot 2^8) : (2^2 \cdot 5^5 \cdot 2^6 \cdot 5^6) = \\\\ = ((2^8 \cdot 5^{11}) \cdot (2^4 \cdot 5 +2 - 5^2)) : (2^8 \cdot 5^{11}) = 2^4 \cdot 5 +2 - 5^2 =\\\\ =16 \cdot 5 +2 - 25 =80+2-25 = 57$

$\frac{5 \cdot 4^{15} \cdot 9^9 - 4 \cdot 3^{20} \cdot 8^9}{5 \cdot 2^9 \cdot 6^{19} - 7\cdot 2^{29} \cdot 27^6} = \frac{5 \cdot (2^2)^{15} \cdot (3^2)^9 - 2^2 \cdot 3^{20} \cdot (2^3)^9}{5 \cdot 2^9 \cdot (2 \cdot 3)^{19} - 7\cdot 2^{29} \cdot (3^3)^6} = \frac{5 \cdot 2^{30} \cdot 3^{18} - 2^2 \cdot 3^{20} \cdot 2^{27}}{5 \cdot 2^9 \cdot 2^{19} \cdot 3^{19} - 7\cdot 2^{29} \cdot 3^{18}} =$

$=\frac{5 \cdot 2^{30} \cdot 3^{18} - 2^{29} \cdot 3^{20}}{5 \cdot 2^{28} \cdot 3^{19} - 7\cdot 2^{29} \cdot 3^{18}} =\frac{(2^{28} \cdot 3^{18}) \cdot (5 \cdot 2^2- 2 \cdot 3^2)}{(2^{28} \cdot 3^{18}) \cdot (5 \cdot 3 - 7\cdot 2 )} =\frac{5 \cdot 2^2- 2 \cdot 3^2}{5 \cdot 3 - 7\cdot 2} = \\\\ =\frac{5 \cdot 4- 2 \cdot 9}{15 - 14} =\frac{20- 18}{1} =2$

Zad. 2

$\frac{7^8 : 49^2 \cdot5^4 +49 \cdot 125 :\frac{1}{5}}{(7 \cdot 5)^4 : 7^4} =\frac{7^8 : (7^2)^2 \cdot5^4 +7^2 \cdot 5^3 \cdot 5}{(7^4 \cdot 5^4) : 7^4} =\frac{7^8 : 7^4 \cdot 5^4 +7^2 \cdot 5^4}{5^4} =\\\\ =\frac{7^4 \cdot 5^4 +7^2 \cdot 5^4}{5^4} =\frac{5^4 \cdot (7^4 +7^2)}{5^4} =7^4 +7^2 = 7^2 \cdot (7^2 +1) = \\\\ = 49 \cdot (49 +1) = 49 \cdot 50 = 2450$

$\frac{216 :6^2 \cdot 9 - 48 :12 \cdot 3^3}{81 - 3 \cdot 18 \cdot 5} + \frac{5}{7} = \frac{6^3 :6^2 \cdot 3^2 - 4 \cdot 3^3}{3^4 - 3 \cdot 3^2 \cdot 2 \cdot 5} + \frac{5}{7} = \frac{6 \cdot 3^2 - 4 \cdot 3^3}{3^4 - 3^3 \cdot 10} + \frac{5}{7} = \\\\ = \frac{3^2 \cdot (6 - 4 \cdot 3)}{3^3 \cdot(3^2 - 3 \cdot 10)} + \frac{5}{7} = \frac{6-12}{9-30} + \frac{5}{7} = \frac{-6}{-21} + \frac{5}{7} =\frac{2}{7} + \frac{5}{7} =\frac{7}{7} = 1$

$2^{10} \cdot (\frac{1}{8})^3 +5^4 \cdot 25^{-1} +64^{-1} \cdot 2^{10} = 2^{10} \cdot \frac{1}{8^3} +5^4 \cdot \frac{1}{25} +\frac{1}{64} \cdot 2^{10} = \\\\ = 2^{10} \cdot \frac{1}{(2^3)^3} +5^4 \cdot \frac{1}{5^2} +\frac{1}{2^6} \cdot 2^{10} = 2^{10} \cdot \frac{1}{2^9} +5^{4-2} +2^{10 -6} = \\\\ 2^{10-9}+5^2 +2^4 = 2 +25 + 16 = 43$

$2^8 \cdot (\frac{1}{2})^3 +2 \cdot 5^4 \cdot (\frac{1}{5})^3 +(4^{-1})^2 \cdot 2 = 2^8 \cdot \frac{1}{2^3} +2 \cdot 5^4 \cdot \frac{1}{5^3} +4^{-2} \cdot 2 = \\\\ = 2^{8-3} +2 \cdot 5^{4-3} +(2^2)^{-2} \cdot 2 =2^5 +2 \cdot 5^1 +2^{-4} \cdot 2 = \\\\ =32 +10 +2^{-4+1}=42 +2^{-3} = 42 +\frac{1}{2^3} = 42 +\frac{1}{8} = 42\frac{1}{8}$