matematica clasa a 6a

lumperdean panait · Mesaj de **lumperdean panait** » 17 Sep 2014, 12:51

determinati numarul natural n ; 49^n :7^n+7(n+2):7=8*4^13

tata2008 · Mesaj de **tata2008** » 17 Sep 2014, 15:42

lumperdean panait scrie:determinati numarul natural n ; 49^n :7^n+7^(n+2):7=8*4^13

$\rm{\bl\\ (1) ... folosim formula: a^n : b^n = (a:b)^n ,(a^m)^n=a^{m\cdot{n}} si a^{m+n}=a^m\cdot{a^n} \\ 49^n=(7^2)^n = 7^{2n}\surd 49^n:7^n=(49:7)^n=7^n \surd \\ 7^{n+2}:7 = 7^{n+2-1} = 7^{n+1} = 7^n\cdot7 \surd \\ (2) ... \underbrace{49^n :7^n}+\underbrace{7^{n+2}:7}=8*4^{13} \Leftrightarrow \underbrace{7^n +7^n\cdot7}_{7^n \cdot 8}=8\cdot4^{13} \Leftrightarrow 7^n \cdot 8 = 8 \cdot 4^{13} | _{:8}\\ 7^n = 4^{13}$

$\rm{\re\Large\\ Oare in loc de 4^{13} nu avem 7^{13} ? \\ La postarea anterioar\breve{a} ai avut 8 \cdot 7^{11} ! \\ Repar\breve{a} greseala si vei avea in final 7^n = 7^{11} adic\breve{a} \fbox{n=11}$

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matematica clasa a 6a

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