Question

Ajutor ex 6,7,8 vă rog dau si coroana

Answer 1

   
[tex]\displaystyle\\ 6)\\ 6^2:[4^2-(5\cdot 3^5:15-5^0):2^3]=\\ =6^2:[4^2-(5\cdot3\cdot 3^4:15-5^0):2^3]=\\ =6^2:[4^2-(15\cdot 3^4:15-5^0):2^3]=\\ =6^2:[4^2-(3^4-5^0):2^3]=\\ =6^2:[4^2-(81-1):8]=\\ =6^2:[4^2-80:8]=\\ =6^2:[16-10]=6^2:6= \boxed{6} [/tex]


[tex]\displaystyle\\ 7)\\ ~[(3^3+7\cdot 3^7:63):3^3-7^0]\cdot 10^2=\\ =[(3^3+7\cdot 3^2\cdot 3^5:63):3^3-7^0]\cdot 10^2=\\ =[(3^3+7\cdot 9\cdot 3^5:63):3^3-7^0]\cdot 10^2=\\ =[(3^3+63\cdot 3^5:63):3^3-7^0]\cdot 10^2=\\ =[(3^3+3^5):3^3-7^0]\cdot 10^2=\\ =[3^3(1+3^2):3^3-7^0]\cdot 10^2=\\ =[(1+3^2)-7^0]\cdot 10^2=\\ =[(1+9)-1]\cdot 10^2=[10-1]\cdot 10^2=9\cdot 100=\boxed{900} [/tex]


[tex]\displaystyle\\ 8)\\ ~[3^4-(2\cdot 2^2\cdot 2^3)^{100}:(2^{150})^4]:4^2=\\ =\Big[3^4-\Big(2^{1+2+3}\Big)^{100}:\Big(2^{150}\Big)^4\Big]:4^2=\\ =\Big[3^4-\Big(2^{6}\Big)^{100}:\Big(2^{150}\Big)^4\Big]:4^2=\\ =\Big[3^4-2^{6\times 100}}:2^{150\times 4}\Big]:4^2=\\ =\Big[3^4-2^{600}}:2^{600}\Big]:4^2=\\ =[3^4-1]:4^2=[81-1]:4^2=80:4^2=80:16 = \boxed{5}[/tex]