Question

express each of the following as rational numbers​

Answer 1

GIVEN :-

$\implies \bf \bigg \{ \bigg( \dfrac{4}{3} \bigg) ^{ - 1} - \bigg( \dfrac{1}{4} \bigg) ^{ - 1} \bigg \} ^{ - 1}$

TO FIND :-

$\implies \bf \: the \: value \: of \: \bigg \{ \bigg( \dfrac{4}{3} \bigg) ^{ - 1} - \bigg( \dfrac{1}{4} \bigg) ^{ - 1} \bigg \} ^{ - 1}$

SOLUTION :-

$\implies \bf \bigg \{ \bigg( \dfrac{4}{3} \bigg) ^{ - 1} - \bigg( \dfrac{1}{4} \bigg) ^{ - 1} \bigg \} ^{ - 1} \\ \\ \implies \bf \bigg \{ \bigg( \dfrac{3}{4} \bigg) - \bigg( \dfrac{4}{1}\bigg) \bigg \} ^{ - 1} \\ \\ \implies \bf \bigg \{ \bigg( \dfrac{3}{4}\bigg) \: - \bigg( \dfrac{4 \times 4}{1 \times 4} \bigg) \bigg \} ^{ - 1} \\ \\ \implies \bf \bigg \{ \bigg( \dfrac{3 - 16}{4} \bigg) \bigg \} ^{ - 1} \\ \\ \implies \bf \bigg \{ \dfrac{ - 13}{4} \bigg \} ^{ - 1} \\ \\ \implies \boxed{ \red{ \bf \bigg \{ \dfrac{ - 4}{13} \bigg \}}}$

Hence the required answer is-4/13.

ADDITIONAL INFORMATION :-

$\boxed{\begin{minipage}{7cm} \\ \sf{ \implies \bf \sqrt[n]{ \sqrt[m]{ \sqrt[p]{((a^{x} )^{y}) ^{z} } } } = (a ^{xyz} )^{ \frac{1}{mnp} } = a ^{ \frac{xyz}{mnp} } } \\ \\ \sf{ \implies a^m \times a^n = a^{m+n} } \\ \\ \sf{ \implies {a}^{m} \times b^m = ab^m } \\ \\ \sf{ \implies \dfrac{a^m}{a^n} = a^{m - n} ( \tt{ If \: m > n} ) } \\ \\ \sf{ \implies \dfrac{a^m}{ a^n} = \dfrac{ 1}{ a^{n-m} } ( \tt{ If \: n > m )} } \\ \\ \sf{ \implies (a^m)^n = a^{mn} } \\ \\ \sf{ \implies a^{-n} = \dfrac{1}{ a^n} }\end{minipage}}$

Answer 2

QUESTION:-

✯.EXPRESS EACH OF THE FOLLOWING AS RATIONAL NUMBERS.

$\star \sf \bigg \{ \bigg[ \dfrac{4}{3} \bigg] ^{ - 1} - \bigg[ \dfrac{1}{4} \bigg] ^{ - 1} \bigg \} ^{ - 1}$

ANSWER✔

$\large\underline\bold{GIVEN,}$

$\therefore \sf \bigg \{ \bigg[ \dfrac{4}{3} \bigg] ^{ - 1} - \bigg[ \dfrac{1}{4} \bigg] ^{ - 1} \bigg \} ^{ - 1}$

$\large\underline\bold{TO\:FIND,}$

$\sf\dashrightarrow value\:(express)\:of\:the\:given\:rational\:number,$

✯. IDENTITY IN USE,

$\sf\dashrightarrow a^{-1}= \dfrac{1}{a}$

$\large\underline\bold{SOLUTION,}$

$\sf\implies \bigg \{ \bigg( \dfrac{4}{3} \bigg) ^{ - 1} - \bigg( \dfrac{1}{4} \bigg) ^{ - 1} \bigg \} ^{ - 1}$

$\sf\implies \bigg \{ \bigg( \dfrac{3}{4} \bigg) - \bigg( \dfrac{1}{4} \bigg) ^{ - 1} \bigg \} ^{ - 1}$

$\sf\implies \bigg \{ \bigg( \dfrac{3}{4} \bigg) - \big( 4\big) \bigg \} ^{ - 1}$

$\sf\implies \bigg \{ \bigg( \dfrac{3}{4} \bigg) - \bigg( \dfrac{4}{1}\bigg) \bigg \} ^{ - 1}$

$\sf\implies \bigg \{ \bigg( \dfrac{3}{4}\bigg) - \bigg( \dfrac{4 \times 4}{1 \times 4} \bigg) \bigg \} ^{ - 1}$

$\sf\implies \bigg \{ \bigg( \dfrac{3 - 16}{4} \bigg) \bigg \} ^{ - 1}$

$\sf\therefore a^{-1}= \dfrac{1}{a}$

$\sf\implies \bigg \{ \dfrac{ - 4}{13} \bigg \}$

$\large{\boxed{\bf{\implies \bigg( \dfrac{- 4}{13} \bigg) }}}$

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✯ADDITIONAL INFORMATION,

FEW IDENTITIES TO LEARN,

$\sf\therefore (a+b)^2=a^2+2ab+b^2$

$\sf\therefore (a+b)(a-b)=a^2-b^2$

$\sf\therefore (a+b)^2+(a-b)^2=2(a^2+b^2)$

$\sf\therefore (a+b)^3=a^3+b^3+3ab(a+b)$

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