Question

ma puteți ajuta cu aceste 4 exerciții, va rog! Promit sa ma abonez la cine imi răspunde. ​

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

 

$\displaystyle\bf\\45.b)\\\\\left[\Big|3^{12}-5^{8}\Big|+(-25)^4\right]\cdot(-2)^{12}-36^6\\\\Ne~ocupam~de~modul.\\3^{12}=3^{3\times4}=(3^3)^4=27^4\\5^8=5^{2\times4}=(5^2)^4=25^4\\27^4>25^4~~\implies~~3^{12}>5^8\\\\\implies~~\Big|3^{12}-5^{8}\Big|=3^{12}-5^8\\\\Ne~ocupam~de~puterile~cu~baza~negativa~si~exponent~par.\\(-25^4)=25^4~deoarece~exponentul~este~numar~par.\\(-2^{12})=2^{12}~deoarece~exponentul~este~numar~par.$

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$\displaystyle\bf\\\left[\Big|3^{12}-5^{8}\Big|+(-25)^4\right]\cdot(-2)^{12}-36^6=\\\\=\left[3^{12}-5^{8}+25^4\right]\cdot2^{12}-36^6=\\\\=\left[3^{12}-5^{8}+\Big(5^2\Big)^4\right]\cdot2^{12}-\Big(6^2\Big)^6=\\\\=\left[3^{12}-5^{8}+5^{2\times4}\right]\cdot2^{12}-\Big(6^{2\times6}\Big)^6=\\\\=\left[3^{12}\underbrace{-5^{8}+5^8}_{=~0}\right]\cdot2^{12}-6^{12}=\\\\=3^{12}\times2^{12}-6^{12}=\\\\=(3\times2)^{12}-6^{12}=\\\\=6^{12}-6^{12}=\boxed{\bf0}$

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$\displaystyle\bf\\45.c)\\\\4^{15}-3\cdot4^{14}-3\cdot4^{13}-3\cdot4^{12}-...-3\cdot4^3-3\cdot4^2-3\cdot4=\\\\=4^{15}-3\Big(4^{14}+4^{13}+4^{12}+...+4^3+4^2+4\Big)=\\\\=4^{15}-3\Big(4+4^2+4^3+...+4^{12}+4^{13}+4^{14}\Big)=\\\\=4^{15}-3\Big(\frac{4^{14+1}-1}{4-1}-1 \Big)=\\\\=4^{15}-3\Big(\frac{4^{15}-1}{3}-1 \Big)=\\\\=4^{15}-3\times\frac{4^{15}-1}{3}+3=\\\\=4^{15}-(4^{15}-1)+3=\\\\=4^{15}-4^{15}+1+3=\\\\=0+1+3=\boxed{\bf4}$

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$\displaystyle\bf\\45.d)\\\\5^{100}-4\cdot5^{99}-4\cdot5^{98}-...-4\cdot5^2-4\cdot5=\\\\=5^{100}-4\Big(5^{99}+5^{98}+...+5^2+5\Big)=\\\\=5^{100}-4\Big(5+5^2+...+5^{98}+5^{99}\Big)=\\\\=5^{100}-4\Big(\frac{5^{99+1}}{5-1}-1 \Big)=\\\\=5^{100}-4\Big(\frac{5^{100}-1}{4}-1 \Big)=\\\\=5^{100}-4\times\frac{5^{100}-1}{4}+4=\\\\=5^{100}-(5^{100}-1)+4=\\\\=5^{100}-5^{100}+1+4=\\\\0+1+4=\boxed{\bf5}$

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$\displaystyle\bf\\\\46.a)\\\\(-5-10-15-...-400):(-1-2-3-...-80)=\\\\=\frac{(-5-10-15-...-400)}{-(-1-2-3-...-80)}=\\\\La~numarator~dam~factor~comun~pe~5\\si ramanem~cu~paranteze~identice~la~numarator\\pe~care~le~simplificam\\\\=\frac{5(-1-2-3-...-80)}{(-1-2-3-...-80)}= \boxed{\bf5}$