Question

simplify:
[(2)-¹ + (4)-¹ +(3)-¹]-¹​

Answer 1

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

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

/* By Exponential Law */

$\boxed{ \pink{a^{-n} = \frac{1}{a^{n}} }}$

$= \Big( \frac{6+3+4}{12}\Big)^{-1}$

$= \Big( \frac{13}{12}\Big)^{-1}$

$= \frac{12}{13}$

Therefore.,

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

$\green{ = \frac{12}{13}}$

•••♪

Answer 2

Step-by-step explanation:

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

−1

+4

−1

+3

−1

)

−1

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

2

1

+

4

1

+

3

1

)

−1

/* By Exponential Law */

\boxed{ \pink{a^{-n} = \frac{1}{a^{n}} }}

a

−n

=

a

n

1

= \Big( \frac{6+3+4}{12}\Big)^{-1}=(

12

6+3+4

)

−1

= \Big( \frac{13}{12}\Big)^{-1}=(

12

13

)

−1

= \frac{12}{13}=

13

12

Therefore.,

\red{ \Big( 2^{-1} + 4^{-1} + 3^{-1} \Big)^{-1}}(2

−1

+4

−1

+3

−1

)

−1

\green{ = \frac{12}{13}}=

13

12

•••♪