Question

1) 336:($7^{2}$+$0^{7}$-$5^{2}$)
2) b$b^{13}$*b*$b ^{21}$*$b^{20}$
3) $16^{4}$*$4 ^{10}$*$64 ^{5}$
4) $c ^{58}$:$c ^{39}$:$c ^{9}$
5) $125^{42}$:$25^{63}$
6) ($8 ^{37}$*$81 ^{25}$):($16 ^{27}$*$27 ^{33}$)
7) $2 ^{1}$*$2 ^{2}$*$2 ^{3}$*$2 ^{4}$*...*$2 ^{45}$
10) (1+$5 ^{10}$):(1+$5 ^{10}$)

Answer 1

$1).336:(7^2+0^7-5^2)=336:(49+0-25)=336:24=14 \\ 2).b^1^3 \cdot b \cdot b^2^1 \cdot b^2^0=b^{13+1+21+20}=b^5^5 \\ 3).16^4 \cdot 4^1^0 \cdot 64^5=(4^2)^4 \cdot 4^1^0 \cdot (4^3)^5=4^8 \cdot 4^1^0 \cdot 4^1^5=4^{8+10+15}=4^3^3 \\ 4).c^5^8:c^3^9:c^9=c^{58-39-9}=c^1^0 \\ 5).125^4^2:25^6^3=(5^3)^4^2:(5^2)^6^3=5^1^2^6:5^1^2^6=1 \\ 6).(8^3^7 \cdot 81^2^5):(16^2^7 \cdot 27^3^3)=[(2^3)^3^7 \cdot (3^4)^2^5]:[(2^4)^2^7 \cdot (3^3)^3^3]= \\ =(2^1^1^1 \cdot 3^1^0^0):(2^1^0^8 \cdot 3^9^9)=2^3 \cdot 3=8 \cdot 3=24$
$\displaystyle 7).2^1 \cdot 2^2 \cdot 2^3 \cdot 2^4 \cdot ... \cdot 2^4^5=1+2+3+...+45= \\ = \frac{45(45+1)}{2} = \frac{45 \cdot 46}{2} = \frac{2070}{2} =1035=2^{1035} \\ 8).(1+5^1^0):(1+5^1^0)=1$