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De considera numarul natural n=(7×7^3×7^4)^6÷(343)^2^4 Sa se calculeze n^2019.​

Răspuns :

[tex]\displaystyle\\n=\Big(7\times7^3\times7^4\Big)^6:(343)^{\b2^4}\\\\n=\frac{\Big(7\times7^3\times7^4\Big)^6}{\Big(343\Big)^{\b2^4}}\\\\n=\frac{\Big(7^{1+3+4}\Big)^6}{\Big(343\Big)^{\b2^4}}=\frac{\Big(7^{8}\Big)^6}{\Big(7^3\Big)^{16}}=\frac{7^{8\times6}}{7^{3\times16}}=\frac{7^{48}}{7^{48}}=1\\\\\boxed{\bf~n=1}\\\\n^{2019}=1^{2019}=\boxed{\bf1}[/tex]