Question

$(\frac{1}{4}{)}^{\frac{-1}{2}}-3(8{)}^{\frac{2}{3}}+(\frac{9}{16}{)}^{\frac{-1}{2}}$

Answer 1

we know,

$\frac{a^{-m}}{b^{-m}}=(\frac{a}{b}{)}^{-m}=(\frac{b}{a}{)}^m$

so question becomes,

$(\frac{4}{1}{)}^{\frac{1}{2}}-3(8{)}^{\frac{2}{3}}+(\frac{16}{9}{)}^{\frac{1}{2}}$

$⟹$

$(2^2{)}^{\frac{1}{2}}-3(2^3{)}^{\frac{2}{3}}+(\frac{4^2}{3^2}{)}^{\frac{1}{2}}$

also,

$(a^m{)}^n=a^{mn}$

so question becomes,

$2-3\ast 4+\frac{4}{3}$

$=-26/3$