Find the value of each of the following:

Solution:

(i) 

$\left(\frac{1}{2}\right)^{-1}+\left(\frac{1}{3}\right)^{-1}+\left(\frac{1}{4}\right)^{-1}=\frac{1}{1 / 2}+\frac{1}{1 / 3}+\frac{1}{1 / 4} \quad \rightarrow>\left(a^{-1}=1 / a\right)$

=2+3+4

 =12

(ii) $\left(\frac{1}{2}\right)^{-2}+\left(\frac{1}{3}\right)^{-2}+14-2=11 / 22+11 / 32+11 / 42 \quad->\left(a^{-n}=1 /\left(a^{n}\right)\right)$

$=\frac{1}{1 / 4}+\frac{1}{1 / 9}+\frac{1}{1 / 16} \quad->\left((a / b)^{n}=\left(a^{n} / b^{n}\right)\right)$ 

= 4+9+16

=29

(iii)

$\left(2^{-1} \times 4^{-1}\right) \div 2^{-2}=\left(\frac{1}{2} \times \frac{1}{4}\right) \div \frac{1}{2^{2}} \quad->\left(a^{-n}=1 /\left(a^{n}\right)\right)$

$=\frac{1}{8} \times 4$

$=2$

(iv) 

$\left(5^{-1} \times 2^{-1}\right) \div 6^{-1}=\left(\frac{1}{5} \times \frac{1}{2}\right) \div \frac{1}{6} \quad->\left(a^{-n}=1 /\left(a^{n}\right)\right)$

$=\frac{1}{10} \times 6$

$=\frac{3}{5}$