Question

${\left(\frac{1}{64}\right)}^0+{\left(64\right)}^{\frac{-1}{2}}+{\left(32\right)}^{\frac{4}{5}}-{\left(32\right)}^{\frac{-4}{5}}$ is equal to

• A. $16\frac{1}{8}$
• B. $17\frac{1}{8}$
• C. $17\frac{1}{16}$
• D. $-17\frac{1}{16}$

Answer 1

The correct option is C $17\frac{1}{16}$

${\left(\frac{1}{64}\right)}^0+{\left(64\right)}^{\frac{-1}{2}}+{\left(32\right)}^{\frac{4}{5}}-{\left(32\right)}^{\frac{-4}{5}}$

$=1+(8^2{)}^{-\frac{1}{2}}+(2^5{)}^{\frac{4}{5}}-(2^5{)}^{-\frac{4}{5}}$

$=1+8^{-1}+2^4-2^{-4}$

$=1+\frac{1}{8}+16-\frac{1}{16}=\frac{16+2+256-1}{16}$

$=\frac{273}{16}$

$=17\frac{1}{16}$