Question

1)Calculați:
A)(25^4)^3:(4^2+3^2)^11-(0^2010+1^2012+2013^0)=
B)[123+2^7*2^2+2^67:2^27]:[123+2^14:2^5+(2^5)^8]=
C)(2^8+9^2+5^6+36^2):(0^6+4^4+3^4+25^3+6^4)=
D)(4+8+12+16+...+8000):(2+4+6+8+....+4000)+(2^53+2^52+2^51):(2^17)^3
E){637+15*[15*(282:47+516:43):45-6]:21}:13=
F)(6+12+18+24+...+600)^2:(3+6+9+12+...+300):(1+2+3+...+100)=

2)Calculați:
A)a=[a^5+a^7+(a^1)^20]^0:[(a^2)^4:(a^4)^2]=
B)b=[(2^3)^5+25^3-7^35:7^20]:(2^15-7^15+5^6)*3^36=
C)c=[2+2^2*2^22+2^200:2^120+(3^3)^10]:[1+2^24+2^80+2^2013+3^30]-2+(5^7)^3=

Please, help me!!Mulțumesc anticipat!

Answer 1

$1a).(25^4)^3:(4^2+3^2)^1^1-(0^2^0^1^0+1^2^0^1^2+2013^0)= \\ =25^1^2:(16+9)^1^1-(0+1+1)=25^1^2:25^1^1-2= \\ =25^1^2^-^1^1-2=25^1-2=25-2=23$
$b).[123+2^7*2^2+2^6^7:2^2^7]:[123+2^1^4:2^5+(2^5)^8]= \\ =(123+2^7^+^2^+^6^7^-^2^7):(123+2^1^4^-^5+2^4^0)= \\ =(123+2^4^9):(123+2^9+2^4^0)=(123+2^4^9):(123+2^9^+^4^0)= \\ =(123+2^4^9):(123+2^4^9)=1$
$c).(2^8+9^2+5^6+36^2):(0^6+4^4+3^4+25^3+6^4)= \\ =[2^8+(3^2)^2+5^6+(6^2)^2]:[0+(2^2)^4+3^4+(5^2)^3+6^4]= \\ =(2^8+3^4+5^6+6^4):(2^8+3^4+5^6+6^4)=1$
$d).(4+8+12+16+...+8000):(2+4+6+8+....+4000)+ \\ +(2^5^3+2^5^2+2^5^1):(2^1^7)^3 \\ 4+8+13+16+...+8000=4(1+2+3+...+2000) \\ 2+4+6+8+...+4000=2(1+2+3+...+2000) \\ 4(1+2+3+...+2000):2(1+2+3+...+2000)=2 \\ (2^5^3+2^5^2+2^5^1):(2^1^7)^3=2^5^1(2^2+2+1):2^5^1= \\ =2^5^1(4+2+1):2^5^1=4+2+1=7 \\ 2+7=9$
e).{637+15*[15*(282:47+516:43):45-6]:21}:13=
={637+15*[15*(6+12):45-6]:21}:13=
=[637+15*(15*18:45-6):21]:13=
=[637+15*(270:45-6):21]:13=
=[637+15*(6-6):21]:13=
=(637+15*0:21):13=
=637:13=
=49
$f).(6+12+18+24+...+600)^2:(3+6+9+12+...+300): \\ :(1+2+3+...+100) \\ (6+12+18+24+...+600)^2=[6(1+2+3+...+100)]^2= \\ = 36(1+2+3+...+100)^2= \\ =36(1+2+3+...+100)(1+2+3+...+100) \\ 3+6+9+12+...+300=3(1+2+3+...+100) \\ 36:3=12$
$2a).a=[a^5+a^7+(a^1)^2^0]^0:[(a^2)^4:(a^4)^2] \\ a=1:(a^8:a^8) \\ a=1:1 \\ a=1$
$b).b=[(2^3)^5+25^3-7^3^5:7^2^0]:(2^1^5-7^1^5+5^6)*3^3^6 \\ b=[2^1^5+(5^2)^3-7^3^5^-^2^0]:(2^1^5-7^1^5+5^6)*3^3^6 \\ b=(2^1^5+5^6-7^1^5):(2^1^5-7^1^5+5^ 6)*3^3^6 \\ b=3^3^6$
$c).[2+2^2*2^2^2+2^2^0^0:2^1^2^0+(3^3)^1^0]:[1+2^2^4+2^8^0+2^2^0^1^ 3+3^3^0]- \\ -2+(5^7)^3 \\ c=(2+2^2^4+2^8^0+3^3^0):(1+2^2^4+2^8^0+2^2^0^1^3+3^3^0)-2+5^2^1 \\ c=1-2+5^2^1 \\ c=5^2^1-1$