Question

(2^-1-3^-1)^1+ (6^-1 - 8^-1)^1​

Answer 1

We know that,

• $\displaystyle\sf {a^{-1}=\dfrac {1}{a}}$

Thus we have,

• $\displaystyle\sf {2^{-1}=\dfrac {1}{2}}$

• $\displaystyle\sf {3^{-1}=\dfrac {1}{3}}$

• $\displaystyle\sf {6^{-1}=\dfrac {1}{6}}$

• $\displaystyle\sf {8^{-1}=\dfrac {1}{8}}$

So,

$\displaystyle\sf{\longrightarrow(2^{-1}-3^{-1})^{-1}+(6^{-1}-8^{-1})^{-1}=\left (\dfrac {1}{2}-\dfrac {1}{3}\right)^{-1}+\left (\dfrac {1}{6}-\dfrac {1}{8}\right)^{-1}}$

$\displaystyle\sf{\longrightarrow(2^{-1}-3^{-1})^{-1}+(6^{-1}-8^{-1})^{-1}=\left (\dfrac {3-2}{2\times 3}\right)^{-1}+\left (\dfrac {8-6}{6\times8}\right)^{-1}}$

$\displaystyle\sf{\longrightarrow(2^{-1}-3^{-1})^{-1}+(6^{-1}-8^{-1})^{-1}=\left (\dfrac {1}{6}\right)^{-1}+\left (\dfrac {2}{48}\right)^{-1}}$

$\displaystyle\sf{\longrightarrow(2^{-1}-3^{-1})^{-1}+(6^{-1}-8^{-1})^{-1}=\left (\dfrac {1}{6}\right)^{-1}+\left (\dfrac {1}{24}\right)^{-1}}$

$\displaystyle\sf{\longrightarrow(2^{-1}-3^{-1})^{-1}+(6^{-1}-8^{-1})^{-1}=6+24}$

$\displaystyle\sf{\longrightarrow\underline {\underline {(2^{-1}-3^{-1})^{-1}+(6^{-1}-8^{-1})^{-1}=30}}}$

Hence 30 is the answer.

Answer 2

Answer:

30 is the correct answer