Question

va rog mult de tot ajuati-ma si pe mine la ex: 3,4,5,6 va rog dau coroana si mulțumiri. e urgent va roggggg !!

Answer 1

   
[tex]\displaystyle\\ 3a)\\ 12^{55}:6^{55}:2^{50} -7^1\cdot 5^0+13^2=\\ =(12^{55}:6^{55}):2^{50} -7^1\cdot 5^0+13^2=\\ =(12:6)^{55}:2^{50} -7^1\cdot 5^0+13^2=\\ =2^{55}:2^{50} -7^1\cdot 5^0+13^2=\\ =2^{55-50} -7^1\cdot 5^0+13^2=\\ =2^{5} -7\cdot 1+13^2=\\ =32 -7+169=\\ =25+169= \boxed{\bf 194}\\\\ 3b)\\ (2\cdot 5)^3+45^{45}:9^{45}:5^{44}-2000^0= =(10)^3+(45:9)^{45}:5^{44}-1=\\ =(10)^3+5^{45}:5^{44}-1=\\ =(10)^3+5^{45-44}-1=\\ =(10)^3+5^{1}-1= 1000+5-1=\boxed{\bf 1004}[/tex]


[tex]\displaystyle\\ 3c)\\ 15^{38}:5^{38}-\Big(3^{19}\Big)^2=\\ =(15:5)^{38} - 3^{19\times2}=\\ =3^{38} - 3^{38}=\boxed{\bf0}\\\\ 3d)\\ =30^{n+3}:10^{n+3}-3^n\cdot 3^3 =\\ =(30:10)^{n+3} - 3^{n+3}=\\ =(3)^{n+3} - 3^{n+3}=\boxed{\bf0} [/tex]


[tex]\displaystyle\\ 4)\\ (4^{50}+4^{49}+4^{48}):21 =\\ =(4^{48+2}+4^{48+1}+4^{48}):21 =\\ =(4^{48}\cdot 4^2+4^{48}\cdot 4^1+4^{48}):21 =\\ =4^{48}\Big(4^2+4^1+1\Big):21 =\\ =4^{48}\Big(16+4+1\Big):21 =\\ =4^{48} \times 21:21 = \boxed{4^{48}}\\ 4^{48} = 4^{2\times 24} = \Big(4^{24}\Big)^2 = \texttt{\bf patrat perfect}\\ 4^{48} = 4^{3\times 16} = \Big(4^{16}\Big)^3 = \texttt{\bf cub perfect}\\[/tex]


[tex]\displaystyle\\ 5)\\ U(S) = U\Big[ U(8^{283})+U(9^{126})\Big]=\\\\ = U\Big[ U(8^{280+3})+U(9^{2\times 63})\Big]=\\\\ = U\Big[ U(8^{280} \times8^3})+U(9^{2\times 63})\Big]=\\\\ = U\Big[ U(8^{4\times 70} \times8^3})+U(9^{2\times 63})\Big]=\\\\ = U\Big[ U\Big(\Big(8^4\Big)^{70} \times8^3}\Big)+U\Big(\Big(9^2\Big)^{63}\Big)\Big]=\\\\ = U\Big[ U\Big(4096^{70} \times 512}\Big) +U\Big(81^{63}\Big)\Big]=\\\\ = U\Big[ U\Big(6 \times 2}\Big) +U\Big(1^{63}\Big)\Big]=\\\\ = U[ U(12+1)]=\boxed{\bf 3}[/tex]


[tex]\displaystyle\\ 6)\\ ~[(x+360:4)\cdot 5+700]:600 = 2\\ ~[(x+90)\cdot 5+700] = 2\times 600\\ (x+90)\cdot 5+700 = 1200\\ (x+90)\cdot 5 = 1200-700\\ (x+90)\cdot 5 = 500\\ x+90 = 500:5\\ x+90 = 100\\ x = 100 - 90 = \boxed{\bf10} [/tex]