Simplify:
(i) $ \left\{4^{-1} \times 3^{-1}\right\}^{2} $
(ii) $ \left\{5^{-1}\div6^{-1}\right\}^{3} $
(iii) $ \left(2^{-1}+3^{-1}\right)^{-1} $
(iv) $ \left\{3^{-1} \times 4^{-1}\right\}^{-1} \times 5^{-1} $
(v) $ \left(4^{-1}-5^{-1}\right)^{-1}\div3^{-1} $



To do:

We have to simplify the given expressions.

Solution:

We know that,

$a^{-m}=\frac{1}{a^m}$ 

Therefore,

(i) $(4^{-1} \times 3^{-1})^{2}=(\frac{1}{4} \times \frac{1}{3})^{2}$

$=(\frac{1}{12})^{2}$

$=\frac{1}{12} \times \frac{1}{12}$

$=\frac{1}{144}$ 

(ii) $(5^{-1} \div 6^{-1})^{3}=(\frac{1}{5} \div \frac{1}{6})^{3}$

$=(\frac{1}{5} \times \frac{6}{1})^{3}$

$=(\frac{6}{5})^{3}$

$=\frac{6 \times 6 \times 6}{5 \times 5 \times 5}$

$=\frac{216}{125}$

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

$=(\frac{3+2}{6})^{-1}$

$=(\frac{5}{6})^{-1}$

$=\frac{6}{5}$

(iv) $(3^{-1} \times 4^{-1})^{-1} \times 5^{-1}=(\frac{1}{3} \times \frac{1}{4})^{-1} \times \frac{1}{5}$

$=(\frac{1}{12})^{-1} \times \frac{1}{5}$

$=\frac{12}{1} \times \frac{1}{5}$

$=\frac{12}{5}$

(v) $(4^{-1}-5^{-1})^{-1}\div3^{-1}=(\frac{1}{4}-\frac{1}{5})^{-1} \div \frac{1}{3}$

$=(\frac{5-4}{20})^{-1} \times \frac{3}{1}$

$=\frac{20}{1} \times \frac{3}{1}$

$=60$

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