Question

Pls help I need this ASAP​

Answer 1

Explanation:

`((2*3^(-2))^(3)(5*3^(2))^(2))/(3^(-2)*(5*2)^(2))=(2^(3)*3^(-2*3)*5^(2)*3^(2*2))/(3^(-2)*5^(2)*2^(2))\\\\\\=(2^(3)*3^(-6)*5^(2)*3^(4))/(3^(-2)*5^(2)*2^(2))\\\\\\=2^(3-2)*3^(-6+4-(-2))*5^(2-2) = 2^(1)*3^(-6+4+2)*5^(0)\\\\\\=2*3^(0)*5^(0)=2*1*1 \\\\\\= 2`

`3) 3^(3)(4^(0))^(2)*(3*2)^(-3) * 2^(2)=3^(3) *1 *3^(-3)*2^(-3)*2^(2)\\\\\\= 3^(3-3)*2^(-3+2) = 3^(0)*2^(-1) = 1*2^(-1)\\\\\\=(1)/(2)`

`3)(3^(7)*4^(7)*(2*5)^(-3)*5^(2))/(12^(7)*5^(-1)*2^(-4))\\\\\\=(3^(7)*((2^(2))^(7)*2^(-3)*5^(-3)*5^(2))/((2^(2)*3)^(7)*5^(-1)*2^(-4))\\\\\\=(3^(7)*2^(2*7)*2^(-3)*5^(-3)*5^(2))/(2^(2*7)*3^(7)*5^(-1)*2^(-4))\\\\\\=(3^(7)*2^(14)*2^(-3)*5^(-3)*5^(2))/(2^(14)*3^(7)*5^(-1)*2^(-4))\\\\\\=3^(7-7)*2^(14-3-14+4)*5^(-3+2+1)\\\\=3^(0)*2^(1)*5^(0)\\\\\\=1*2*1\\\\\\=2`

Hint:

`a^(0)=1\\\\(a^(m))/(a^(n))=a^(m-n)\\\\\\a^(m)*a^(n)=a^(m+n)\\\\\\a^(-m)=(1)/(a^(m))`