Question

Calculeaza: (3^202•3^3+3^303:3^98+9^100•3^5):3^205-1^2013

Answer 1

[tex](3^{202}\cdot3^3+3^{303}:3^{98}+9^{100}\cdot3^5):3^{205}-1^{2013}=\\ (3^{205}+3^{205}+3^{205}):3^{205}-1=\\ 3^{205}(1+1+1):3^{205}-1=3\cdot3^{205}:3^{205}-1=3-1=2\\ 9^{100}=(3^2)^{100}=3^{200}\\ 3^{200}\cdot3^5=3^{205}[/tex]

Answer 2

$(3^{202} \cdot 3^3+3^3^0^3:3^9^8+9^1^0^0 \cdot 3^5):3^2^0^5-1^2^0^1^3= \\ =[3^2^0^2^+^3+3^3^0^3^-^9^8+(3^2)^1^0^0 \cdot3^5]:3^{205}-1= \\ =(3^2^0^5+3^2^0^5+3^2^0^0 \cdot 3^5):3^2^0^5-1= \\ =(3^2^0^5+3^2^0^5+3^2^0^0^+^5):3^2^0^5-1=(3^2^0^5+3^2^0^5+3^2^0^5):3^2^0^5-1= \\ =1+1+1-1=2+1-1=3-1=2$