Question

solve the question seventh​

Answer 1

Answer:

Hope answer will help u

Ho

Answer 2

${(x \div y)}^{ - 1} = {x}^{ - 1} \div {y}^{ - 1} \\ {( \dfrac{x}{y}) }^{ - 1} = \frac{ \dfrac{1}{x} }{ \dfrac{1}{y} } \\ \dfrac{y}{x} = \frac{y}{x}$

By taking any value, both are equal.

${(x \div y)}^{ - 1} = {x}^{ - 1} \div {y}^{ - 1} \\ {( \frac{7}{11} \div ( - \frac{22}{21} ))}^{ - 1} = { (\frac{7}{11}) }^{ - 1} \div { (- \frac{22}{21} )}^{ - 1} \\ {( \dfrac{7 \div 11}{ - 22 \div 21}) }^{ - 1} = \frac{ \dfrac{1}{7 \div 11} }{ \dfrac{1}{ - 22 \div 21} } \\ \dfrac{ - 22 \times 11}{7 \times 21} = \frac{- 22 \times 11}{7 \times 21}$

LHS = RHS

★ Hence proved