Question

cu metoda factorului comun pls​

Answer 1

 

$\displaystyle\bf\\a)\\7^{32}+4\times7^{30}=\\=7^{30+2}+4\times7^{30}=\\=7^{30}\times7^2+4\times7^{30}=\\=7^{30}(7^2+4)=7^{30}(49+4)=7^{30}\times53\\7^{30}\times53:53=\boxed{\bf7^{30}}\\\\b)\\5^{23}+5^{22}+5^{21}=\\=5^{21+2}+5^{21+1}+5^{21}=\\=5^{21}\times5^2+5^{21}\times5^1+5^{21}=\\=5^{21}(5^2+5^1+1)=5^{21}(25+5+1)=5^{21}\times31\\5^{21}\times31:31=\boxed{\bf5^{21}}$

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$\displaystyle\bf\\c)\\3^{100}+3^{101}+3^{102}+3^{103}=\\=3^{100}+3^{100+1}+3^{100+2}+3^{100+3}=\\=3^{100}+3^{100}\times3^1+3^{100}\times3^2+3^{100}\times3^3=\\=3^{100}(1+3^1+3^2+3^3)=3^{100}(1+3+9+27)=3^{100}\times40\\3^{100}\times40:40=\boxed{\bf3^{100}}$

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$\displaystyle\bf\\d)\\8^{41}+8^{42}+8^{43}+8^{44}+8^{45}+8^{46}=\\=\Big(8^{41}+8^{41+1}+8^{41+2}\Big)+\Big(8^{44}+8^{44+1}+8^{44+2}\Big)=\\=\Big(8^{41}+8^{41}\times8^1+8^{41}\times8^2\Big)+\Big(8^{44}+8^{44}\times8^1+8^{44}\times8^2\Big)=\\=8^{41}\Big(1+8^1+8^2\Big)+8^{44}\Big(1+8^1+8^2\Big)=\\=8^{41}\Big(1+8+64\Big)+8^{44}\Big(1+8+64\Big)=\\8^{41}\times73+8^{44}\times73=73\Big(8^{41}+8^{44}\Big)\\73\Big(8^{41}+8^{44}\Big):73=\boxed{\bf8^{41}+8^{44}}$