Question

Verify the above Question

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Answer 1

Step-by-step explanation:

${(x + y)}^{ - 1} = {( \frac{2}{3} - \frac{5}{6} )}^{ - 1} \\ \\ = {( \frac{4 - 5}{6}) }^{ - 1} \\ \\ = {( \frac{ - 1}{6} )}^{ - 1} \\ \\ = - 6.$

${x}^{ - 1} + {y}^{ - 1} = {( \frac{2}{3}) }^{ - 1} + {( \frac{ - 5}{6} )}^{ - 1} \\ \\ = \frac{3}{2} - \frac{6}{5} \\ \\ = \frac{15 - 12}{10} \\ \\ = \frac{3}{10} .$

$therefore \: \: {(x + y)}^{ - 1} \: is \: not \: equal \: to \: \: \: \: \: {x}^{ - 1} + {y}^{ - 1} .$