Math, asked by tayyaba1981, 4 months ago

Simplify :
(i)
 \frac{ {81}^{n} . \: 3^{5} -  {(3)}^{4n - 1} (243) }{ ({9}^{2n})( {3}^{3})  }
(ii)
 \frac{4(3)^n}{ {3}^{n + 1}  -  {3}^{n} }

Answers

Answered by BilalFNA
1

ANSWER:

(i). 6

(ii). 2

EXPLANATION:

\frac{ {81}^{n} . \: 3^{5} - {(3)}^{4n - 1} (243) }{ ({9}^{2n})( {3}^{3})}

 =\frac{( {3}^{4})^{n}  . \: 3^{5} - {3}^{4n }.  \: {3}^{ - 1}  . \:  { 3}^{5}  }{ ({ {3}^{3}) }^{2n}.( {3}^{3}) }

 =\frac{{3}^{4n}  . \: 3^{5} - \:  {3}^{4n }.  \: {3}^{4}  }{ { {3}^{4n}}.  \: {3}^{3}}

 =\frac{{3}^{4n}  . \: 3^{5}   }{ { {3}^{4n}}.  \: {3}^{3}} -  \frac{\:  {3}^{4n }.  \: {3}^{4}}{\:  {3}^{4n }.  \: {3}^{3}}

 =  \frac{ {3}^{5}}{ {3}^{3} }  -  \frac{ {3}^{4} }{ {3}^{3} }

 =  {3}^{5 - 3}  -  {3}^{4 - 3}

 =  {3}^{2}  - 3

 = 9 - 3

 = 6

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 \frac{4(3)^n}{ {3}^{n + 1} - {3}^{n} }

 = \frac{4(3^n)}{ {3}^{n}(3 - 1) }

 =  \frac{ {4( {3}^{n}) } }{3^{n}(2) }

 = \frac{4}{2}

 = 2

I hope it helps you. Please mark me as BRAINLIEST if you like.

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