Question

simplify [7^-1+[3/2]^-1]^-1​

Answer 1

Answer :

= ${[{7}^{-1} + [{\dfrac{3}{2}}^{-1}}]^{-1}$

= ${[ \dfrac{1}{7}+ [\dfrac{2}{3}]}^{-1}$

= ${[ \dfrac{3+14}{21}]}^{-1}$

= ${[ \dfrac{17}{21}]}^{-1}$

= $\dfrac{21}{17}$

•°• 21/17 is your required answer !!

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Answer 2

$( {7}^{ - 1} + ( \frac{3}{2} ) {}^{ - 1} ) { }^{ - 1}$

$( \frac{1}{7} + \frac{2}{3} ) {}^{ - 1}$

$( \frac{1}{7} \times \frac{3}{3} + \frac{2}{3} \times \frac{7}{7} ) {}^{ - 1}$

$( \frac{3 + 14}{21} ) {}^{ - 1}$

$( \frac{17}{21} ) {}^{ - 1}$

$\frac{21}{17}$

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