Question

va rog mult am nevoie de ajutor ex 4 si 5

Answer 1

[tex]4.a).(3^0*3^1*3^{13}:3^5+1):[(3^2)^3*27+1) = \\=(1*3*3^{13}:3^5 +1):(3^6*3^3+1)= \\ =(3*3^8+1):(3^9+1)= \\ =(3^9+1):(3^9+1)= \\ =1 \\ b).(2^3+2^3)^2:(2+2^2):2^2+2^0*0^2= \\ =(2*2^3)^2 : (2*2^2):2^2 +1*0= \\ =(2^4)^2:2^3:2^2+0= \\ =2^8:2^3:2^2= \\ =8 \\ c). (3*5^{12}+2*5^{12}):5^{11}= \\ =5*5^{12}:5^{11}= \\ =5^2= \\ =25 \\ d).[(2^5)^3:2^5*(2^4+2^4)-8^4:2^{10}]:(2^{15}-2^2)= \\ =(2^{15}:2^5*2*2^4-2^{12}:2^{10}):[(2^{13}-1)*2^2]= \\ =(2^{15}-2^2):[(8192-1)*4]= \\ =(2^{13}-1)*2^2:(8191*4)= \\[/tex]
$=(8192-1)*4:32764= \\ =8191*4:32764= \\ =32764:32764= \\ =1 \\ e).(2^1-1^2)(3^2-2^3)(4^2-2^4)(2^5-5^2)(2^6-6^2)(2^7-7^2)= \\ =(2-1)(9-8)(2^4-2^4)*(32-25)[2^6-(3*2)^2](2^7-49)= \\ =1*1*0*7(2^6-3^2*2^2)(2^7-49)= \\ =0*7(2^4-3^2)*2^2*(2^7-49)= \\ =0*7(16-9)*4(2^7-49)= \\ =0*7*7*4(2^7-49)= \\ =0*7*28(2^7-49)= \\ =0$
$5.a).2010^{2010}-2009*2010^{2009}-2010^{2009}= \\ =2010^{2010}-2010*2010^{2009}= \\ =2010^{2010}-2010^{2010}= \\ =0 \\ b).(2009+2009^2):(2009^0+2009^1) = \\ =(2009+2009^2):(1:2009) = \\ = (2009+2009^2):2010= \\ \\ = \frac{2009+2009^2}{2010} \\ \\ c).2010^{2010}-2009*2010^{2009}= \\ =(2010-2009)*2010^{2009}= \\ =1*2010^{2009}= \\ =2010^{2009} \\ d).111*888^{2009}+888^{2009}*777-888^{2010}= \\ =(111+777-888)*888^{2009}= \\ =0*888^{2009}= \\ =0 \\ e). (10^{2010}-10^{2009}):(10^{2009}-10^{2008})= \\  =(10-1)*10^{2009}:[(10-1)*10^{2008})= \\ = 9*10^{2009}:(9*10^{2008})= \\ =10 \\ f). 11^4-10*11^3-10*11^2-10*11-10= \\ =14641-10*11^3-10*121-110-10= \\ =14641-10*11^3-1210-110-10= \\ =13311-10*1331= \\ =13311-13310= \\ =1$