Find the values of each of the following
(i) $ 3^{-1}+4^{-1} $
(ii) $ \left(3^{0}+4^{-1}\right) \times 2^{2} $
(iii) $ \left(3^{-1}+4^{-1}+5^{-1}\right)^{0} $
(iv) $ \left\{\left(\frac{1}{3}\right)^{-1}-\left(\frac{1}{4}\right)^{-1}\right\}^{-1} $

To do:  

We have to find the values of each of the  given expressions.

Solution:

We know that,

$a^{-m}=\frac{1}{a^{m}}$

$a^o=1$

Therefore,

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

$=\frac{1\times4+1\times3}{12}$          (LCM of 3 and 4 is 12)

$=\frac{4+3}{12}$

$=\frac{7}{12}$

(ii) $(3^{0}+4^{-1}) \times 2^{2}=(1+\frac{1}{4})\times4$

$=\frac{1\times4+1}{4}\times4$

$=\frac{5}{4}\times4$

$=5$

(iii) $(3^{-1}+4^{-1}+5^{-1})^{0}=(\frac{1}{3}+\frac{1}{4}+\frac{1}{5})^o$

$=1$

(iv) $[{(\frac{1}{3})^{-1}-(\frac{1}{4})^{-1}}]^{-1} =[{(3)^{1}-(4)^{1}}]^{-1}$

$=(3-4)^{-1}$

$=(-1)^{-1}$

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

$=\frac{1}{-1}$

$=-1$

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Updated on: 10-Oct-2022

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