Question

Va rog!!! Am nevoide de ajutor

Answer 1

$a)3^{12}:3^{10}+(3\cdot5)^{10}:15^{9}-2011^{0}\cdot1^{2011}=3^{12-10}+(3\cdot5)^{10}:(3\cdot5)^{9}-1\cdot1=3^{2}+(3\cdot5)^{10-9}-1=9+(3\cdot5)-1=9+15-1=23$

$b)4^{13}:8^{6}+7^{3}-5^{2}\cdot[625:5^{4}+16:(7\cdot3^{2}-55)]=\\ \\2^{26}:2^{3\cdot6}+7^{3}-5^{2}\cdot[5^{4}:5^{4}+2^{4}:(7\cdot9-55)]= \\ \\2^{26-18}+7^{3}-5^{2}\cdot[1+2^{4}:(63-55)]=\\ \\2^{8}+7^{3}-5^{2}\cdot(1+2^{4}:2^{3})= 2^{8}+7^{3}-5^{2}\cdot(1+2)=256+343-75=524$

$c)18\cdot[1+3^{7}\cdot3^{53}+7^{84}:7^{14}+(5^{4})^{20}]:[13^{0}+3^{60}+7^{70}+5^{80}]=$

$18\cdot[1+3^{7+53}+7^{84-14}+(5^{4\cdot20})}]:[1+3^{60}+7^{70}+5^{80}]=\\ \\ 18\cdot[1+3^{60}+7^{70}+5^{80}]:[1+3^{60}+7^{70}+5^{80}]=18$

$8^{n}+8^{n+1}=18\cdot2^{2003};\\ \\ 8^{n}(1+8)=9\cdot2\cdot2^{2003}\\ \\ 8^{n}\cdot9=9\cdot2^{2004}\\ \\ 2^{3n}=2^{2004}\\ \\3n=2004;\\\\ n =2004:3\\ \\n=668$

$9^{n}+9^{n+1}=10\cdot3^{2012}\\ \\9^{n}(1+9)=10\cdot3^{2012}\\ \\9^{n}\cdot10=10\cdot3^{2012}\\ \\3^{2n}=3^{2012}\\ \\2n=2012;\\ \\ n=1006$